SEBA Class 8 Mathematics Chapter 9 বীজগণিতীয় ৰাশি আৰু অভেদসমূহ

SEBA Class 8 Mathematics Chapter 9 বীজগণিতীয় ৰাশি আৰু অভেদসমূহ Question Answer, SEBA Class 8 Maths Notes in Assamese Medium, SEBA Class 8 Maths Solutions in Assamese to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 8 Mathematics Chapter 9 বীজগণিতীয় ৰাশি আৰু অভেদসমূহ Notes and select needs one.

SEBA Class 8 Mathematics Chapter 9 বীজগণিতীয় ৰাশি আৰু অভেদসমূহ

Join Telegram channel

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 8 Mathematics Chapter 9 বীজগণিতীয় ৰাশি আৰু অভেদসমূহ Notes. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 8 Mathematics Chapter 9 বীজগণিতীয় ৰাশি আৰু অভেদসমূহ Solutions for All Subject, You can practice these here.

বীজগণিতীয় ৰাশি আৰু অভেদসমূহ

Chapter – 9

অনুশীলনী – 9.1

1. পূৰণফল উলিওৱাঃ

(i) 3x² × 11xy × 2/3 y²

উত্তৰঃ 3x² × 11xy × 2/3 y²

= 3 × 11 × 2/3 × x × x × y²

= 22x² × x³

= 22x² x³

(ii) (-5x) × 3a² × (-3ax)

WhatsApp Group Join Now
Telegram Group Join Now
Instagram Join Now

উত্তৰঃ (-5x) × 3a² × (-3ax)

= (-5) × 3 × (-3) × x × x × a × a × a

= 45x²a³

(iii) (-3pq) × (-15p³q³) × q²

উত্তৰঃ (-3pq) × (-15p³q³) × q²

= (-3) × (-15) × p x p³ × q × q³ × q²

= 45p⁶ x q⁶

= 45p⁶q⁶

(iv) 3x (5x² + 8)

উত্তৰঃ 3x (5x² + 8)

= 3x × 5x² + 3x × 8

= 3 × 5 × x × x² + 3 × 8 × x

= 15x³ + 24x

(v) 2/3 y(18y²- y)

উত্তৰঃ 2/3  y(18y²- y)

= 2/3 y(18y² – 2/3y × y)

= 2/3 × 18 × y × y² –  2/3 × 1× y × y

= 12y3 – 2/3 y² 

(vi) (-8a³) (a + 3b + 2c)

উত্তৰঃ (-8a³) (a + 3b + 2c) 

= -8a³ × a + (-8a³) × 3b + (-8a³) × 2c 

= -8a⁴ + (-8) × 3 × a³ × b + (-8) × 2 × a³ × c

= -8a⁴ – 24a³b – 16a³c

(vii) (3mn-2n) (-2m²n)

উত্তৰঃ (3mn-2n) (-2m²n) 

= 3mn × (-2m²n) – 2n × (-2m²n)

= 3 × (-2) × m × n × m² × n – 2 × (-2) × n × m² ×n 

= – 6m³n² + 4m²n²

(viii) (9x²+ 4x + 3) × 11x

উত্তৰঃ (9x²+ 4x + 3) × 11x

= 9x² × 11x + 4x × 11x + 3 ×11x 

= 9 × 11 × x² × x + 4 × 11x + 3 × 11x

= 99x³ + 449x² + 33x

(ix) (20a² – 3b² + ab) × (-7b²)

উত্তৰঃ (20a²-3b² + ab) × (-7b²)

= 20a² × (-7b²) – 3b² × (-7b²) + ab × (-7b²) 

= 20 × (-7) × a² × b² – 3 × (-7) × b² × b² + 1 × (-7) ab × b²

= -140a²b² + 21b4 – 7ab³

 = 21b⁴ – 7ab³ – 140a²b²

(x) 3 x³y² (x y + xy³ – 2) 

উত্তৰঃ 3x³y² (xy + xy³ – 2)

= 3x 3y × xy + 3x³y² xxy³ – 3x³ y²x2

= 3x4y³ + 3 x4y5 – 6x³y²

2. পূৰণফল উলিওৱাঃ

(i) (x² + y) (3x²y – y²)

উত্তৰঃ (x² + y) (3x²y – y²)

= x² (3x²y – y²) + y × (3x²y – y²)

= (x² × 3x²y) – x² × y² + (y × 3x²y) – y × y²

= 3x⁴ – x²y² + 3x²y² – y³

= 3x³y + 2x²y²- y³

(ii) (7x – 2y) (2x + 7y)

উত্তৰঃ (7x – 2y) (2x + 7y)

= 7x × (2x + 7y) – 2y × (2x + 7y)

= 7x × 2x + 7x × 7y – 2y × 2x – 2y × 7y

= 14x² + 49xy – 4xy – 14y²

= 14x² + 45xy -14y²

(iii) (1/4 a² + 3b) ⁺ (a³ + 2/3 b²)

উত্তৰঃ (1/4 a² + 3b) (a³ + 2/3 b²)

= 1/4 a² × (a³ + 2/3 b²) + 3b × (a³ + 2/3 b²)

= 1/4 a² × a³ + 1/4 a² × 2/3 b² + 3b × a³ + 3b × 2/3 b²

= 1/4 a⁵ + 2/12 a² b² + 3a³ b + 2b³

= 1/4 a⁵ + 1/6 a² b² + 3a³ b + 2b³

(iv) (1.5x-2.5y) (2.5x – 1.5y) 

উত্তৰঃ (1.5x – 2.5y) (2.5x – 1.5y) 

= 1.5x × (2.5x – 1.5y) – 2.5y × (2.5x – 1.5y) 

= 1.5x × 2.5x – 1.5x × 1.5y – 2.5y × 2.5x + 2.5y × 1.5y

= 3.75x² – 2.25xy – 6.25xy + 3.75y²

= 3.75x²- 8.5xy +3.75y²

(v) (3x + 4y) (2x² + 3y + xy) 

উত্তৰঃ (3x + 4y) (2x² + 3y + xy)

= 3x × (2x² + 3y + xy) + 4y × (2x² + 3y + xy)

= 3x × 2x² + 3x × 3y + 3x × xy + 4y × 2x² + 4y × 3y + 4y ×  xy

= 6x³ + 9xy + 3x²y + 8x²y + 12y² + 4xy² 

= 6x³ + 9xy + 11x²y + 4xy² + 12y²

= 6x³ + 11x²y + 4xy² + 12y² + 9xy

(vi) (2xy + 5x²) (x5y4 – x³y² + xy)

উত্তৰঃ (2xy + 5x²) (x5y4 – x³y² + xy)

= 2xy × (x5y4 – x³y² + xy) + 5x² × (x5y4 – x³y² + xy)

= 2xy × x5y4 – 2xy × x³y² + 2xy × xy + 5x² × x5y4 – 5x² × x³y² + 5x² × xy

= 2x6y5 – 2x4y³ + 2x²y² + 5x7y4 – 5x5y² + 5x³y 

= 2x6y5 – 2x4y³ + 2x²y² + 5x7y4 – 5x5y² + 5x³y

(vii) (3a²b² – 4c) (a³b³ + 2a4b³c³ – 6abc)

উত্তৰঃ (3a²b² – 4c) (a³b³ + 2a4b³c³ – 6abc)

= 3a²b² × (a³b³+2a4b³c³ – 6abc) – 4c × (a³b³ + 2a4b³c³ – 6abc)

= 3a²b²× a³b³ + 3a²b² × 2a4b³c³ – 3a²b² × 6abc – 4c × a³b³- 4c × 2a4b³c3 + 4c × 6abc

= 3a5b5 + 6a6b5c³ – 18a³b³c – 4a³b³c – 8a4b³c4 + 24abc² 

= 3a5b5 + 6a6b5c³ – 22a³b³c – 8a4b³c4 + 24abc²

(viii) (4x²y -5xy² + 3xy) (3x³y – 2)

উত্তৰঃ (4x²y -5xy² + 3xy) (3x³y – 2)

= 4x²y × (3x³y – 2) – 5xy² × (3x³y – 2) + 3xy × (3x³y – 2) 

= 4x²y × 3x³y + 4x²y × (-2) – 5xy² × 3x³y – 5xy² × (-2) + 3xy × 3x³y + 3xy × (-2)

= 12x5y² – 8x²y – 15x4y³ + 10xy² + 9x4y² – 6xy 

= 12x5y² – 15x4y³ + 9x4y² – 8x²y + 10xy² – 6xy

(ix) (2x+3y+z) (5x+2y+1)

উত্তৰঃ (2x+3y+z) (5x+2y+1)

= 2x × (5x + 2y + 1) + 3y × (5x + 2y + 1) + z × (5x + 2y + 1) 

= 2x × 5x + 2x × 2y + 2x × 1 + 3y × 5x + 3y × 2y + 3y × 1 + z × 5x + z × 2y + z × 1

= 10x² + 4xy + 2x + 15xy + 6y² + 3y + 5xz + 2zy + z 

= 10x² + 6y² + 19xy + 2yz + 5zx + 2x + 3y + z

(x) (3x³- 2y² + z) (3x³ + 2y² – z)

উত্তৰঃ (3x³- 2y² + z) (3x³ + 2y²- z)

= 3x³ × (3x³ + 2y²- z) – 2y² × (3x³ + 2y²- z) + z ×(3x³ + 2y²- z)

= 3x³ × 3x³ + 3x³ × 2y² + 3x³ × (-z) – 2y² × 3x³ – 2y² × 2y² – 2y² × (-z) + z × 3×3 + z × 2y2x + z (–z) 

= 9x⁶ – 4y⁴ + 4y²z – z²

3. তলত দিয়া ৰাশিসমূহ সৰল কৰাঃ

(i) 3×(5x+8) – 10x

উত্তৰঃ 3 × (5x + 8) – 10x

= 3x × 5x + 3x × 8 -10x

= 15x² + 24x – 10x

= 15x² + (24-10) x

= 15x² + 14x

(ii) (2m +3m²) (-2mn)

উত্তৰঃ (2m + 3m²) (-2mn)

= 2m × (-2mn) + 3m² × (-2mn)

= – 4m²n – 6m³n 

(iii) 8(3a + 4b) + 5 

উত্তৰঃ 8(3a + 4b) + 5

= 8 × 3a + 8 × 4b + 5

= 24a + 32b +5 

(iv) 2x² (4x-1) + 3 × (x – 3)

উত্তৰঃ 2x²(4x-1) + 3×(x – 3)

= 2x²×4x – 2x²×1 + 3x × x + 3x × (-3)

= 8x³ – 2x² + 3x² – 9x 

= 8x³ + x² – 9x

4. সৰল কৰাঃ

(i) (p + q²) (q²-p)+15 

উত্তৰঃ (p + q²) (q²-p)+15

= p(q²-p) + q(q²-p) + 15 

= pq²- p² + q4 – pq² + 15 

= – p² + q4 + 15 

(ii) (a – b) (a² + ab + b²) + 3b³ 

উত্তৰঃ (a – b) (a² + ab + b²) + 3b³

= a(a² + ab + b²) – b(a² + ab + b²) + 3b³

= a³ + a²b + ab²- a²b – ab²- b³ + 3b³ 

= a³- b³ + 3b³ 

= a³ + 2b³

(iii) y² (y³ + 3x) + y (2xy + y²)

উত্তৰঃ y²(³ + 3x) + y (2xy + y²)

= y2×y³ + y²×3x + y×2xy + y×y²

= y5 + 3xy² + 2xy² + y³

= y³ + 5xy² + y³

(iv) (2/3 x⁴ y³ + 4/9 xy³)1/4-1/6 x⁴y³

উত্তৰঃ (2/3 x⁴ y³ + 4/9 xy³)1/4-1/6 x⁴y³

= 2/3 x³ y³ ×1/4 + 4/9 xy³ × 1/4-1/6 x⁴ y³

= (x⁴ y³)/6 + 1/9 xy³ – 1/6 x⁴ y³

= 1/9 xy³

(v) y³(4y + 5) – (2y + 1) (y³ + 2y² + 1)

উত্তৰঃ y³ (4y + 5) – (2y + 1) (y³ + 2y² + 1) 

= y³ (4y+5) – [2y(y³ + 2y² + 1) + 1(y³ + 2y² + 1)]

= y³ × 4y + y³ × 5 – [2y× y³ + 2y × 2y² + 2y × 1 + y3 + 2y2 + 1] 

= 4y⁴ + 5y³ – 2y⁴ – 4y³ – 2y – y³ – 2y² – 1

= 4y⁴ – 2y⁴ + 5y³ – 4y³ – y³ – 2y² – 2y – 1

= 2y⁴ – 2y² – 2y – 1

(vi) (1.2l – 2.5m) (2.5l + 0.2m+ 1.2) + 0.06l + 7m

উত্তৰঃ (1.2l – 2.5m) (2.5l + 0.2m+ 1.2) + 0.06l + 7m

= 1.2l (2.5l + 0.2m + 1.2) – 2.5m (2.5l + 0.2m + 1.2 ) + 0.06l + 7m 

= 1.2l × 2.5l + 1.2l × 0.2m + 1.2l × 1.2 – 2.5m × 2.5l – 2.5m x 0.2m – 2.5m × 1.2 + 0.06l + 7m

= 3l² + 0.24lm + 1.44l – 6.25lm – 0.5m² – 3m + 0.06l + 7m 

= 3l² + 0.24lm – 6.25lm + 1.44l + 0.06l – 0.5m2 – 3m+7m 

= 3l² + 6.01lm +1.5l – 0.5m² – 3m + 7m

অনুশীলনী 9.2

1. অভেদ (x + a) (x + b) = x2 + (a + b)x + ab ব্যৱহাৰ কৰি তলৰ ৰাশিবোৰ পূৰণ কৰা।

(i) (x + 7) (x + 5)

উত্তৰঃ (x + 7 ) (x + 5) 

= x² + ( 7+ 5) x + 7 × 5

= x² + 12x + 35

(ii) (7x + 2y) (7x + 6y)

উত্তৰঃ (7x + 2y) (7x + 6y) 

= (7x)² + (2y+6y) ×7x+2y×6y

= 49×2 + 8y×7x + 12y2 

= 49×2 + 56xy +12y2

(iii) (4x³ + 8) (4x³ + 10)

উত্তৰঃ (4x³ + 8) (4x³ + 10) 

= (4x³)² + ( 8 + 10).4x³ + 8 × 10

= 16x⁶ + 18 × 4x³ + 80

= 16x⁶ + 72x³ + 80

(iv) (4k² – 3k) (4k² – 7k)

উত্তৰঃ (4k2 – 3k) (4k2 – 7k)

= (4k²) + {(-3k) + (-7k)} 4k² + (-3k) × (-7k)

= 16k⁴ + {–3k-7k} 4k² + 21k²

= 16k⁴ – 10k × 4k² + 21k²

= 16 k⁴ – 40k³ + 21k²

(v) (a/2+1/2)(a/2-1/4)

উত্তৰঃ (a/2 + 1/2)(a/2 – 1/4)

= (a/2)ᶻ + {1/2+(-1/4)(a/2)} +1/2×(-1/4)

= aᶻ/4 + (1/2-1/4) × 9/2 – 1/8

= aᶻ/4 +1/2×a/2 – 1/8

= aᶻ/4 + 4a/8 – 1/8

(vi) (nᶻ/5 – 0.6)(nᶻ/5 + 1.6)

উত্তৰঃ (nᶻ/5 – 0.6)(nᶻ/5 + 1.6)

= (nᶻ/5)² + {(-0.6) + 1.6} (nᶻ/5) + (-0.6) × 1.6

= n⁴/25 + {(-0.6) + 1.6}  nᶻ/5 – 0.96

= n⁴/25 + 1 × nᶻ/5 – 0.96

= n⁴/25 + nᶻ/5 – 0.96

(vii) 98 × 97 

উত্তৰঃ 98 × 97 

= (100 – 2) (100 – 3)

= (100)² + ((-2) + (-3)) (100) + (-2) × (-3)

= 10000 + (-2 – 3) × 100 + 6 

= 10000 – 5 × 100 + 6

= 10000 – 500 + 6

= 10006 – 500

= 9506

(viii) 501 × 503

উত্তৰঃ 501 × 503 

= (500 + 1) (500 + 3) 

= (500)² + (1 + 3) × 500 + 1 × 3

= 250000 + 4 × 500 + 3

= 250000 + 2000 + 3

= 252003

2. অভেদ (a + b)² = a² + 2ab + b² ব্যৱহাৰ কৰি তলৰ বৰ্গবোৰৰ মান উলিওৱা।

(i) (x + 5)²

উত্তৰঃ (x + 5)² 

= x² + 2 × x × 5 + 5²

= x² + 10x + 25

(ii) (5x + 4y)²

উত্তৰঃ (5x + 4y)²

= (5x)² + 2 (5x) (4y) + (4y)²

= 25x² + 40xy  + 16y²

(iii) (3a³ + 4a²)²

উত্তৰঃ (3a³ + 4a²)²

= (3a³)² + 2(3a³) (4a²) + (4a²)²

= 9a⁶ + 24a⁵ + 16а⁴

(iv) (3x – 1/3x)²

উত্তৰঃ (3x – 1/3x)²

= (3x)² + 2 × (3x) × 1/3x + (1/3x)²

= 9x² + 2 × 3x ×  1/3x + 1/9x²

= 9x² + 2 + 1/9x²

= 9x² + 1/9x² + 2

(v) (p/q-q/p)²

উত্তৰঃ (p/q – q/p)²

= (p/q)² + 2 × p/q × q/p + (q/p)²

= p²/q² + 2 + q²/p²

= p²/q² + q²/p² + 2

(vi) (502)²

উত্তৰঃ (502)²

= (500 + 2)²

= (500)² + 2 × 500 × 2 + (2)² 

= 250000 + 2000 + 4

= 252004

(vii) (9.5)2

উত্তৰঃ (9.5)2

= (9 + 0.5)2

= (9)² + 2 × 9 × 0.5 + (0.5)²

= 81 + 9.0 + 0.25

= 90.25

(viii) (4 1/8)²

উত্তৰঃ (4 1/8)²

= 4² + 2 × 4 × 1/8 + (1/8)²

= 16 + 1 + 1/64

= 17+ 1/64

= 17 1/64

3. অভেদ (a – b)2 = a2 – 2ab + b2  ব্যৱহাৰ কৰি তলৰ বৰ্গবোৰৰ মান উলিওৱা।

(i) (x-7)²

উত্তৰঃ (x-7)²

= x² – 2 × x × 7 × 7²

= x² – 14x + 49

(ii) (6x – 5)²

উত্তৰঃ (6x – 5)²

= (6x)² – 2 × 6x × 5 + (5)² 

= 36x² – 60x + 25

(iii) (10x² – 3y)2

উত্তৰঃ (10x² -3y)²

= (10x²)² – 2 × 10x² × 3y + (3y)²

= 100×4 – 60x2y + 9y²

(iv) (p² – q²)

উত্তৰঃ (p² – q²)

= (p²)² – 2 p²q² + (q2)² 

= p⁴ – 2p²q² + q⁴

(v) (a²x – ax²)³ 

উত্তৰঃ (a²x – ax²)³

= (a²x)²- 2 × a²x × ax² + (ax²)² 

= a⁴x² –  2a³x³ + a²x⁴

(vi) (x² – 1/x²)²

উত্তৰঃ (x² -1/x² )²

= (x²)² – 2 × x² ×1/x²  + (1/x² )²

= x⁴ – 2 + 1/x⁴

= x⁴ + 1/x⁴ – 2

(vii) 296²

উত্তৰঃ (296)² 

= (300 – 4)²

= (300)² – 2 × 300 × 4 + 4²

= 90000 – 2400 +16 

= 90000 + 16 – 2400

= 90016 – 2400

= 87616

(viii) (1999)²

উত্তৰঃ (1999)²

= (2000 – 1)²

= (2000) – 2 × 2000 × 1 + 1²

= 4000000 – 4000 + 1

= 4000000 + 1 – 4000

= 4000001 – 4000

= 3996001

4. অভেদ (a + b) (a – b) = a² – b²  ব্যৱহাৰ কৰি তলৰ ৰাশিবোৰ পূৰ্ণ কৰা।

(i) (y + 11) (y – 11)

উত্তৰঃ (y +11) (y –11) 

= y² – (11)²

= y² – 121

(ii) (2x + 3) (2x – 3)

উত্তৰঃ (2x + 3) (2x – 3) 

= (2x)² – 3²

= 4x² – 9

(iii) (6 + m²) (6m² – 6) [বিকল্প ( 6 + m²) (m²– 6)]

উত্তৰঃ (6 + m²) (6m² – 6)

= (m2 + 6) (m2 – 6)

= (m²)² – 6²

= m⁴ – 36

(iv) (ax² – by) (ax² + by)

উত্তৰঃ (ax² – by) (ax² + by)

= (ax² + by) (ax² – by)

= (ax²)² – (by)²

= a²x⁴ – b²y²

(v) (1 – xm) (1 + xm)

উত্তৰঃ (1 – xᵐ) (1 + xᵐ)

= 1² – (xᵐ)²

= 1 – x²ᵐ

(vi) 61 × 59

উত্তৰঃ 61× 59

= (60 + 1) (60 – 1)

= (60)² – 1² 

= 3600 – 1

= 3599

(vii) 106 × 94

উত্তৰঃ 106 × 94 

= (100 + 6) (100 – 6)

= (100)²- (6)²

= 10000 – 36

= 9964

(viii) 9.5 × 8.5

উত্তৰঃ 9.5 × 8.5 

= (9 + 0.5 ) (9 – 0.5)

= (9)² – (0.5)² 

= 81 – 0.25

= 80.75

5. উপযুক্ত অভেদ ব্যৱহাৰ কৰি তলৰ ৰাশিবোৰৰ পূৰণফল উলিওৱা।

(i) (3x – 5m) (3x – 5m)

উত্তৰঃ (3x – 5m) (3x – 5m)

= (3x – 5m)² [(a – b)² = a² – 2ab + b²] 

= (3x)² – 2 × 3x × 5m + (5m)²

= 9x² – 30xm + 25m²

(ii) (4m + 3) (4m + 2)

উত্তৰঃ (4m + 3) (4m + 2) 

= (4m)² + (3 + 2) × 4m + 3 × 2 [(x + a) (x + b) = x² + (a + b)x + ab]

= 16m² + 5 × 4m + 6 

= 16m² + 20m + 6

(iii) (9 + 4n)²

উত্তৰঃ (9 + 4n)2

= 9² + 2 × 9 × 4n + (4n)² [(a+b)² = a² + 2ab + b²]

= 81 + 72n + 16n² 

= 16n² + 72n +81

(iv) (6x + 1/3) (6x + 3) 

উত্তৰঃ (6x + 1/3)(6x + 3)

= (6x)² + (1/3+ 3)6x +1/3×3 [(x+a)(x+b)=x²+(a+b)x+ab]

= 36x² + 10/3 × 6x + 1 

= 36x² + 20x + 1

(v) (4ab – c) (4ab + c)

উত্তৰঃ (4ab – c) (4ab + c)

= (4ab + c) (4ab – c) [(a + b) (a-b) = a² – b²]

= (4ab)² – c² 

= 16a²b² – c²

(vi) (x-x/2)2

উত্তৰঃ (x – x/2)²

= (x)² – 2 × x × x/2 + (x/2)² [(a+b)² = a² – 2ab + b²]

= x² – x² + x²/4

= x²/4

(vii) (a²/2 + b²/4) (a²/2 + b²/4)

উত্তৰঃ (a²/2 + b²/4) (a²/2 + b²/4)

= (a²/2 + b²/4)²

= (a²/2)² + 2 × a²/2 × b²/4 + (b²/4)²

= a⁴/4 + (a² b²)/4 + b⁴/16

(viii) (0.5x² – 0.2y²)² 

উত্তৰঃ (0.5x² – 0.2y²)²

= (0.5x²)² – 2(0.5x²) (0.2y²) + (0.2y²)²

= 0.25×4 – 0.2x²y² + 0.04y² [(a-b)² = a² – 2ab + b²]

(ix) (-9x² + y3) (9x² + y³)

উত্তৰঃ (-9x² + y3) (9x² + y³)

= (9x² + y³) (9x² + y3

= (y³- 9x²) (y³ + 9x²) [(a+b)(a – b)=a²- b²]

= (y³ + 9x²) (y³- 9x²)

= (y³)² – (9x²)²

= y⁶ – 81x⁴

(x) (y²/2 – 4) (y²/2 + 6)

উত্তৰঃ (y²/2 – 4) (y²/2 + 6) [(x + a)(x + b) = x² + (a + b)x + ab]

= (y²/2)² + {-4 + 6}  y²/2 + (-4) × 6

= y⁴/4 + 2 × y²/2 – 24

= y⁴/4 + y² – 24

(xi) (7x²+1/3)2

উত্তৰঃ (7x²+1/3)2

=(7x²)² + 2 × 7x² × 1/3  + (1/3)² [(a + b)² = a² + 2ab + b2]

= 49x⁴ + 14/3 x² + 1/9

(xii) (x + y + z) (x + y – z) 

উত্তৰঃ (x + y + z) (x + y – z)

[(a+b)(a-b)=a²- b²] 

আৰু (a+b)²=a²+2ab+b²]

= {(x+y)+z} {(x+y)-z}

= (x + y)² – z²

= x² + 2xy + y²- z²

= x² + y² – z² + 2xy

(xiii) 1002 × 999 

উত্তৰঃ 1002 × 999

= (1000 + 2) (1000 – 1)

= (1000)² + {2+(-1)} × 1000 + 2 × (-1) 

= 1000000 + 1000 – 2

= 1001000 – 2

= 1000998

(xiv) (10.2)²

উত্তৰঃ (10.2)²

= (10 + 0.2)² [(a + b)² = a² + 2ab + b²]

= (10)² + 2 × 10 × 0.2 + (0.2)²

= 100 + 4 + 0.04

= 104 + 0.04

= 104.04

(xv) 792

উত্তৰঃ 792 

= (80 – 1)² [(a-b)2=a2-2ab+b2]

= (80)2 – 2 × 80 × 1 + 12

= 6400 – 160 + 1

= 6401 – 160

= 6241

(xvi) 6.2 × 5.8

উত্তৰঃ 6.2 × 5.8

= (6 + 0.2) (6 – 0.2) [(a + b) (a – b) = a²- b²]

= (6)² – (0.2)²

= 36 – 0.04

= 35.96

6. সৰল কৰাঃ

(i) (x + 1/x)(2x + 1/x)

উত্তৰঃ (x + 1/x)(2x + 1/x)

= x(2x + 1/x) + 1/x (2x + 1/x)

= x × 2x + x ×1/x + 2x + 1/x × 1/x

= 2 x² + 1 + 2 + 1/x²

= 2 x² +3 + 1/x²

(ii) (2l + m)² – (2l – m)²

উত্তৰঃ (2l + m)² – (2l – m)²

= {(2l + m) + (2l – m)} {(2l + m) – (21 – m)} [a-b² = (a + b)(a – b]

= (2l + m + 21 – m) (2l + m – 2l + m)

= (2l + 2l) (m + m) 

= 4l × 2m

= 8lm

(iii) (a²b + ab²)² – 6a³b³

উত্তৰঃ (a²b + ab²)² – 6a³b³

= (a²b)² + 2a²b × ab² + (ab²)² – 6a³b³

= a4b² + 2a³b³ + a²b4 – 6a³b³

= a4b² + 2a³b³ – 6a³b³ + a²b4

= a4b² – 4a³b³ + a²b4

(iv) (x + y) (x – y) + (y + z) (y – z) + (z + x) (z – x)

উত্তৰঃ (x + y) (x – y) + (y + z) (y – z) + (z + x) (z – x)

= x² – y² + y²- z² + z²- x² 

= x²- x² + y² – y² + z² – z²

= 0

(v) (5a – 6b)² + 20ab – (6b+ 5a)²

উত্তৰঃ (5a – 6b)² + 20ab – (6b+ 5a)²

= (5a – 6b)²- (6b + 5a)² + 20ab

= {(5a – 6b) + (6b + 5a)} {(5a – 6b) – (6b + 5a)} + 20ab

= (5a – 6b + 6b + 5a) (5a – 6b – 6b – 5a) + 20ab

= (5a + 5a) (- 6b – 6b) + 20ab

= 10a × (-12b) + 20ab

=-120ab + 20ab

= -100ab

(vi) (4p² + 5q²) (4p² – 5q2) + (2p² – 5q²)²

উত্তৰঃ (4p² + 5q²) (4p² – 5q2) + (2p² – 5q²)²

= (4p²)² – (5q²)² + (2p²)2 – 2 × 2p² × 5q² + (5q²)² 

= 16p⁴ – 25q⁴ + 4p⁴ – 2p²q² + 25q⁴

= 16p⁴ + 4p⁴ – 25q⁴ + 25q⁴ – 20p²q²

= 20p⁴ – 25q⁴ + 25q⁴ – 20p²q²

= 20p⁴ – 20p²q²

(vii) (2x – 5) (2x + 3) – (x – 2)² + 29

উত্তৰঃ (2x – 5) (2x + 3) – (x – 2)² + 29

= (2x)² + (-5+3) × 2x + (-5) x 3 – (x² -22x+2²)+29

= 4x²+(-2) × 2x – 15 – x² + 4x – 4 + 29

= 4x² –  4x -15 – x² + 4x – 4 + 29

= 4x² – x² – 4x + 4x +29 – 15 – 4

= 3x² + 29 – 19

= 3x² + 10

(viii) (x/3 – 3y/4)(x/3 + 3y/4) + (3y/4 + 3x)(3y/4 + 3x)

উত্তৰঃ (x/3 – 3y/4)(x/3 + 3y/4) + (3y/4 + 3x) (3y/4 + 3x)

= (x/3)² – (2y/4)² + 9y²/16 + 3xy/4 + 9xy/4 + 3x²

= x²/9 – 9y²/16 + 9y²/16 + 3xy/4 + 9xy/4 + 3x²

= x²/9 +3x² + 3xy/4 + 9xy/4

= x² + 27x²)/9 + (3xy + 9xy)/4

= 28x²/9 + 12xy/4

= 28x²/9 + 3xy

(ix) (x/5 + y/5)² + 2(x/5 + y/5)(x/5 – y/5) + (x/5 – y/5)²

উত্তৰঃ (x/5 + y/5)² + 2(x/5 + y/5) (x/5 – y/5) + (x/5 – y/5)²

যদি, ধৰা হয়, x/5 + y/5 = a আৰু 

x/5 – y/5 = b

∴ প্রদত্ত ৰাশিটো হ’ব,

a² + 2ab + b²

= (a + b)²

= {(x/5 + y/5) + (x/5 – y/5)}²

= (x/5 + y/5 + x/5 – y/5)²

= (x/5 + x/5)²

= (2 × x/5)²

= 4/25 x²

(x) 2.89 × 2.89 + 0.22 × 2.89 + 0.0121

উত্তৰঃ 2.89 × 2.89 + 0.22 × 2.89 + 0.0121

[প্রদত্ত ৰাশিটো a2+2ab+b2 আকাৰলৈ ৰূপান্তৰ কৰি]

= 2.89 × 2.89 + 2 × 0.11 × 2.89 + 0.11 

= (2.89)² + 2× (0.11) × 2.89 + (0.11)²

= (2.89 + 0.11)²

= (3)²

= 9

(xi) (0.25 – 2 × 0.5 × 3.5 + 12.25)/3

উত্তৰঃ (0.25 – 2 × 0.5 × 3.5 + 12.25)/3

= (0.25 – 3.5 + 12.25)/3

= (12.5 – 3.5)/3

= 9/3

= 3

(xii) (4.68 × 4.68 – 3.32 × 3.32)/1.36

উত্তৰঃ (4.68 × 4.68 – 3.32 × 3.32)/1.36

= (4.68)² – (3.32)²/1.36

= (4.68+3.32) – (4.68 – 3.32)/1.36

= (8 × 1.36)/1.36

= 8

7. দেখুওৱা যে,

(i) (a-b + c–d)² – (a + b – c + d)² + 4a (b + d) = 4ca

উত্তৰঃ বাওঁপক্ষ = (a – b + c – d)² – (a + b – c + d)2 + 4a (b + d)

= {(a – b + c – d) + (a + b – c + d)} {(a – b + c – d) – (a + b – c + d)} + 4a(b + d) 

= (a – b + c – d + a + b – c + d) (a – b + c – d – a – b + c – d) + 4a(b + d)

= 2a × (-2b + 2c – 2d) + 4a(b + d)

= – 4ab + 4ac – 4ad + 4ab + 4ad 

= 4ac

= সোপক্ষ (দেখুওৱা হ’ল)

(ii) (1.5x² + 1.2y)² – (1.5×2 – 1.2y)² = 7.2x²y

উত্তৰঃ বাওঁপক্ষ = (1.5x² + 1.2y)² – (1.5x² – 1.2y)²

= {(1.5x² + 1.2y) + (1.5x²–1.2y)} {(1.5x² + 1.2y) – (1.5x² – 1.2y)} 

= (1.5x² + 1.2y + 1.5x² – 1.2y) (1.5x² + 1.2y – 1.5x² + 1.2y)

= (1.5x² + 1.5x²) (1.2y + 1.2y)

= 3x² × 2.4y

= 7.2 x²y

= সোপক্ষ (দেখুওৱা হ’ল)

(iii) (2/3 x² + 5)² – 25 =  4/9 x⁴ + 20/3 x²

উত্তৰঃ বাওঁপক্ষ = (2/3 x² + 5)² – 25

= (2/3 x² + 5)² – 5²

= (2/3 x² + 5 + 5)(2/3 x² + 5 – 5)

= (2/3 x² + 10) × 2/3 x²

= 2/3 x² × 2/3 x² + 10 × 2/3 x²

= 4/9 x⁴ + 20/3 x² 

= সোপক্ষ (প্রমাণিত)

(iv) (a² + b²) (a² – b²) + (b² + c²) (b² – c²) + 2c² (c² – a²)

উত্তৰঃ বাওঁপক্ষ = (a² + b²) (a² – b²) + (b² + c²) (b² – c²) + 2c² (c² – a²)

= (a²)² – (b²)² + (b²)² – (c²)² + 2c² (c² – a²) 

= a⁴ + b⁴ +  b⁴ – c⁴ + 2c² (c² – a²)

= a⁴ – c⁴ + 2c² (c² – a²)

= a⁴ – c⁴ + 2c⁴ -2c²a²

= a⁴ + c⁴ -2c²a²

= (c²)² – 2c²a² + (a²)²

=(c² – a²)²

= সোপক্ষ (প্রমাণিত)

8. উপযুক্ত অভেদ ব্যৱহাৰ কৰি তলৰ সমস্যাবোৰ সমাধান কৰা।

(i) এখন আয়তাকৃতি পথাৰৰ দীঘ (x + 8) মিটাৰ প্ৰস্থ দীঘতকৈ 3 মিটাৰ কম। খেল পথাৰখনৰ কালি উলিওৱা।

উত্তৰঃ দিয়া আছে, 

আয়তাকৃতি পথাৰৰ দীঘ = (x + 8) মিটাৰ

প্রস্থ = (x + 8 – 3) মিটাৰ 

= (x + 5) মিটাৰ

∴ পথাৰখনৰ কালি = দীঘ × প্রস্থ

= (x + 8 ) (x + 5 ) বর্গমিটাৰ।

= x² + (8 + 5 ) x + 8 × 5 বর্গমিটাৰ

= x² + 13x + 40 বর্গমিটাৰ

∴ নিৰ্ণেয় পথাৰখনৰ কালি = x² + 13x + 40 বর্গমিটাৰ

(ii) এখন বর্গাকৃতি বাগিচাৰ দীঘ (2x + 1/4 ) মিটাৰ। বাগিচাখনৰ কালি নিৰ্ণয় কৰা।

উত্তৰঃ দিয়া আছে, 

বর্গাকৃতি বাগিচাৰ দীঘ কালি = (2x + 1/4 ) মিটাৰ

বাগিচাখনৰ কালি = (বাহু)²

= (2x + 1/4)² বৰ্গ মিটাৰ

= (2x)² + 2 × 2x × 1/4 + (1/4)²

= 4x² + x + 1/16  বৰ্গ মিটাৰ।

∴ নিৰ্ণেয় বাগিচাখনৰ কালি = 4x² + x + 1/16 বৰ্গ মিটাৰ।

(iii) এজন খেতিয়কে পথাৰত দুডোখৰ বৰ্গাকৃতিৰ মাটিত জহাধান আৰু বৰাধানৰ খেতি কৰিলে। জহা ধানৰ খেতি কৰা মাটি ডৰাৰ দীঘ বৰাধানৰ খেতি কৰা মাটি ডৰাতকৈ 5 মিটাৰ বেচি। দুয়ো ডৰা মাটিৰ কালিব পার্থক্য উলিওৱা।

উত্তৰঃ ধৰাহ’ল,

বৰাধানৰ খেতি কৰা মাটি ডৰাৰ দীঘ মিটাৰ = x মিটাৰ।

∴ জহাধানৰ খেতি কৰা মাটি ডৰাৰ দীঘ = (x + 5) মিটাৰ। 

∴ দুয়োডৰা মাটিৰ কালি ক্ৰমে x² আৰু (x + 5)² বর্গমিটাৰ ৷

∴ পার্থক্য = (x + 5)² – x² 

= x² + 2 × x × 5 + 52 – x²

= (10x + 25) বর্গমিটাৰ

[∵ (x + 5)> x]

(iv) যদি 1 বর্গমিটাৰ বেৰত ৰং কৰোতে 9.00 টকা খৰচ হয়, তেন্তে 107 মিটাৰ দীঘ আৰু 93 মিটাৰ প্ৰস্থৰ বেৰ এখন ৰং কৰিবলৈ কিমান টকাৰ প্ৰয়োজন হ’ব। 

উত্তৰঃ  দিয়া আছে,

বেৰখনৰ দীঘ = 107 মিটাৰ আৰু

বেৰখনৰ প্ৰস্থ = 93 মিটাৰ

∴ বেৰখনৰ কালি = দীঘ × প্রস্থ

= 107 × 93 বর্গমিটাৰ। 

= (100 + 7) (100–7) বর্গ মিটাৰ

= (100) 2 – (7)2 বর্গ মিটাৰ

= (10000–49) বর্গমিটাৰ 

= 9951 বর্গমিটাৰ

∴ ৰং দিবলগীয়া ঠাইৰ পৰিমাণ = 9951 বর্গমিটাৰ।

এতিয়া, 1 বর্গমিটাৰত ৰং দিয়া খৰছ = 9.00 টকা। 

∴ 9951 বর্গমিটাৰত ৰং দিয়া খৰছ = (9 × 9951) টকা

= 89,559.00 টকা।

∴ নিৰ্ণেয় বেৰত ৰং দিয়া মুঠ খৰছ = 89,559.00 টকা।

(v) 197 মিটাৰ দৈৰ্ঘ্যৰ বৰ্গাকৃতি পথাৰ এখনৰ কালি উলিওৱা। 

উত্তৰঃ বর্গাকৃতি পথাৰখনৰ দৈৰ্ঘ্য = 197 মিটাৰ

প্ৰস্থ = 197 মিটাৰ

∴ পথাৰখনৰ কালি = দীঘ × প্রস্থ

= 197 × 197

= (200-3) (200-3) 

= (200-3)²

= (200)² -2 × 200 × 3 +3² 

= 40000 – 1200 + 9

= 40009 – 1200

= 38,809 বর্গমিটাৰ।

∴ নিৰ্ণেয় পথাৰখনৰ কালি = 38,809 বর্গমিটাৰ।

(vi) x + 1/x  = 3 হলে, x² + 1/x²  আৰু x⁴ + 1/x⁴  ৰ মান নির্ণয় কৰা। 

উত্তৰঃ দিয়া আছে, 

x + 1/x = 3

⇒ (x + 1/x)² = 3² [দুয়োফালে বৰ্গ কৰি]

⇒ x² + (1/x)² + 2.x.1/x = 9

⇒ x² + 1/x² + 2 = 9

⇒ x² + 1/x² = 9 – 2

⇒ x² + 1/x² = 7

∴ x² + 1/x² = 7

আকৌ, x² + 1/x² = 7

⇒ (x² + 1/x²)² = 7² [দুয়োফালে বৰ্গ কৰি]

⇒ (x²)² + 2x² (1/x²) + (1/x²)² = 49

⇒ x⁴ + 2 + 1/x² + 2 = 49 

⇒ x⁴ + 1/x⁴ = 49 – 2

⇒ x⁴ + 1/x⁴ = 47

∴ x⁴ + 1/x⁴ = 47

[বিকল্প নিয়মঃ

x² + 1/x²  = x² + (1/x)² + 2x 1/x – 2x 1/x

= (x+1/x )² – 2 

= 3² – 2 

= 9 – 2 

= 7]

(vii) 2x + 1/2x  = 2 হলে, 4x² +1/4x² আৰু 16x⁴ -1/16x⁴  ৰ মান নির্ণয় কৰা।

উত্তৰঃ দিয়া আছে,

2x + 1/2x = 2

⇒ (x² + 1/x²)² = 2² [দুয়োফালে বৰ্গ কৰি]

⇒ (2x)² – 2 2x² (1/2x) + (1/2x)² = 4

⇒ 4x² – 2 + 1/4x² = 4 

⇒ 4x² + 1/4x² = 4 + 2

⇒ 4x² + 1/4x² = 6

∴ 4x² +1/4x² = 6

আকৌ,

4x² + 1/4x = 6

⇒ (4x² + 1/4x²)² = 6² [দুয়োফালে বৰ্গ কৰি]

⇒ (4x²)² – 24x² (1/4x²) + (1/4x²)² = 36

⇒ 16x² – 2 + 1/16x² = 36

⇒ 16x² + 1/16x² = 36 – 2

⇒ 16x² + 1/16x² = 34

∴ 16x² + 1/16x² = 34

(viii) a – b = 10 আৰু ab = 11 হ’লে, a + b ৰ মান নির্ণয় কৰা। 

উত্তৰঃ দিয়া আছে, 

ab = 10 আৰু 

ab = 11 হ’লে,

a + b ৰ মান নির্ণয় কৰিব লাগে- 

(a + b)² – (a – b)² = 4ab 

⇒ (a + b)² = (ab)² + 4ab

= 10² + 4 x 11

= 100 + 44

= 144

⇒ (a + b)² = (12)²

∴ a + b = 12

(a+b)² = a² + 2ab + b²

(a – b)² = a² – 2ab + b²

∴ (a + b)² – (a – b)² = 4ab

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top