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SEBA Class 8 Mathematics Chapter 9 বীজগণিতীয় ৰাশি আৰু অভেদসমূহ
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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)
উত্তৰঃ (-5x) × 3a² × (-3ax)
= (-5) × 3 × (-3) × x × x × a2 × a
= 45x²a³
(iii) (-3pq) × (-15p³q³) × q²
উত্তৰঃ (-3pq) × (-15p³q³) × q²
= (-3) × (-15) × p x p³ × q × q³ × q²
= 45p4 x q⁶
= 45p4q⁶
(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 × 11 × x × x + 3 × 11 × x
= 99x³ + 44x² + 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)
= 3x3 3y2 × xy + 3x³y² × xy³ + 3x³ y² × (-2)
= 3 ×1 × x3 × y2 × x × y + 3 × 1 × x3 × y2 × x y3 + 3 × (-2) × x3 ×y2
= 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
= 3x⁴y – 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
= 7× 2 × x × x × 7 × 7 × x × y – 2 × 2 × y × x – 2 × 7 × y × y
= 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 × (-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 × 3x3 + z × 2y2 + 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) + 3x (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
= y5 + 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
= 49x2 + 8y× 7x + 12y²
= 49x² + 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)2 + {1/2+(-1/4)(a/2)} + 1/2 × (-1/4)
= a²/4 + (1/2-1/4) × 9/2 – 1/8
= a²/4 +1/2 × a/2 – 1/8
= a²/4 + 4a/8 – 1/8
(vi) (n²/5 – 0.6)(nᶻ/5 + 1.6)
উত্তৰঃ (n²/5 – 0.6)(n²/5 + 1.6)
= (n²/5)² + {(-0.6) + 1.6} (n²/5) + (-0.6) × 1.6
= n⁴/25 + {(-0.6) + 1.6} n²/5 – 0.96
= n⁴/25 + 1 × n²/5 – 0.96
= n⁴/25 + n²/5 – 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 + 2x (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
= 100x⁴ – 60x²y + 9y²
(iv) (p² – q²)
উত্তৰঃ (p² – q²)
= (p²)² – 2 p²q² + (q2)²
= p⁴ – 2p²q² + q⁴
(v) (a²x – ax²)²
উত্তৰঃ (a²x – ax²)²
= (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)
= (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
= 0.25x4 – 0.2x²y² + 0.04y4 [(a-b)² = a² – 2ab + b²]
(ix) (-9x² + y3) (9x² + y³)
উত্তৰঃ (-9x² + y3) (9x² + y³) [(a+b) (a – b) = a²- b²]
= (y³- 9x²) (y³ + 9x²)
= (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 [(a + b) (a – b) = a²- b²]
= (6 + 0.2) (b – 0.2)
= (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)} [a2-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 – (1.5x² – 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
= c4 – 2c2a2 + a4
= (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/x⁴ = 49
⇒ x⁴ + 1/x⁴ = 49 – 2
⇒ 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
⇒ (2x – 1/2x)2 = 22
⇒ (2x)2 + (1/2x)2 – 2.2x. 1/2x =4
⇒ 4x² + 1/4x2 – 2 = 4
⇒ 4x² + 1/4x² = 4 + 2
⇒ 4x² + 1/4x² = 6
আকৌ,
4x² + 1/4x² = 6
⇒ (4x² + 1/4x²)² = 6²
⇒ (4x2)2 + (1/4x2)2 + 2.4x2 × 1/4x2 = 36
⇒ 16x4 + 1/16x² = 36 – 2
⇒ 16x4 + 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 = (a – b)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
