8类RD Sharma解决方案–第6章代数表达和恒等式–练习6.4 |套装2

第6章代数表达式和恒等式–练习6.4 |套装1

问题11.找到1.5x(10x ² y – 100xy ² )的乘积

解决方案：

Using Distributive law,

1.5x (10x²y – 100xy²) = (1.5x) × (10x²y) – (1.5x) × (100xy²)

= 15x²⁺¹y – 150x¹⁺¹y² = 15x³y – 150x²y²

Hence, the product is 15x³y – 150x²y²

问题12.查找4.1xy(1.1xy)的乘积

解决方案：

Using Distributive law,

4.1xy (1.1x-y) = 4.1xy × 1.1x – 4.1xy × y

= 4.51x¹⁺¹y – 4.1xy¹⁺¹

= 4.51x²y – 4.1xy²

Hence, the product is 4.51x²y – 4.1xy²

问题13：找出250.5xy(xz + y / 10)的乘积

解决方案：

Using Distributive law,

250.5xy (xz + y/10) = 250.5x¹⁺¹yz + 25.05xy¹⁺¹

= 250.5x²yz + 25.05xz²

Hence, the product is 250.5x²yz + 25.05xz²

问题14.查找7X ² y的乘积/ ⁵⁽² 3XY / 5 + 2×/ 5)

解决方案：

Using Distributive law,

7x²y/5 (3xy²/5 + 2x/5) = 21x²⁺¹y¹⁺²/25 + 14x²⁺¹y/25

= 21x³y³/25 + 14x³y/25

Hence, the product is 21x³y³/25 + 14x³y/25

问题15.查找4a / 3的乘积(a ² + b ² -3c ² )

解决方案：

Using Distributive law,

4a/3 (a² + b² -3c²) = 4a¹⁺²/3 + 4ab²/3 – 4ac²

= 4a³/3 + 4ab²/3 – 4ac²

Hence, the product is 4a³/3 + 4ab²/3 – 4ac²

问题16。找到乘积24x ² (1 – 2x)并评估x = 3的乘积。

解决方案：

Using Distributive law,

24x² (1 – 2x) = 24x² – 48x³

The product is 24x² – 48x³

Now put x = 3

So, 24(3)² – 48(3)³

= 216 – 1296 = -1080

The answer came out to be -1080

问题17。找到-3y(xy + y ² )的乘积，并找到x = 4和y = 5的乘积。

解决方案：

Using Distributive law,

-3y (xy +y²) = -3xy² – 3y³

The product is -3xy² – 3y³

Now put x = 4, and y = 5

So, -3(4)(5)² – 3(5)³

= -300 – 375 = -675

The answer came out to be -675

问题18乘法- 3 X ² Y ^3/2(2× – y)和验证对于x = 1且y = 2的答案

解决方案：

Using Distributive law,

– 3 x²y³/2(2x – y) = -3x³y³ + 3x²y⁴/2

The product is -3x³y³ + 3x²y⁴/2

Now put x = 1, and y = 2 and verifying the L.H.S and R.H.S

L.H.S = – 3 x²y³/2 (2x – y)

= -3(1)²(2)³/2 [2(1) – 2]

= 0

R.H.S = -3x³y³ + 3x²y⁴/2

= -3(1³)(2)³ + 3(1)²(2)⁴/2 = -24 + 24 

= 0 

Since, L.H.S = R.H.S = 0

Hence verified

问题19将二项式乘以二项式，求出x = -1，y = 0.25和z = 0.05的值：
(i)15y ² (2 – 3x)
(ii)-3x(y ² + z ² )
(iii)z ² (x – y)
(iv)xz(x ² + y ² )

解决方案：

i)15岁² (2 – 3x)

Using Distributive law,

15y² (2 – 3x) = 30y² – 45xy²

The product of given monomial and binomial is 30y² – 45xy²

Now put x = -1 and y = 0.25 

So, 30(0.25)² – 45(-1)(0.25)²

= 1.8750 + 2.8125

= 4.6875

The answer will be 4.6875

ii)-3x(y ² + z ² )

Using Distributive law,

-3x (y² + z²) = -3xy² – 3xz²

The product of given monomial and binomial is -3xy² – 3xz²

Now put x = -1, y = 0.25 and z =0.05

-3(-1)(0.25)² – 3(-1)(0.05)²

= 0.1875 + 0.0075 

= 0.1950

The answer will be 0.1950

iii)z ² (x – y)

Using Distributive law,

z²(x – y) = z²x – z²y

The product of given monomial and binomial is z²x – z²y

Now put x = -1, y = 0.25 and z =0.05

= (0.05)²(-1) – (0.05)²(0.25)

= -0.0025 – 0.000625

= -0.003125

The answer will be -0.003125

iv)xz(x ² + y ² )

Using Distributive law,

xz(x² + y²) = x³z + xy²z

The product of given monomial and binomial is x³z + xy²z

Now put x = -1, y = 0.25 and z =0.05

= (-1)³(0.05) + (-1)(0.25)²(0.05)

= -0.05 – 0.003125

= -0.053125

The answer will be -0.053125

问题20简化：
(i)2x ² (在¹ – x处)– 3x(x ⁴ + 2x)-2(x ⁴ – 3x ² )
(ii)x ³ y(x ² – 2x)+ 2xy(x ³ – x ⁴ )
(iii)3a ² + 2(a + 2)– 3a(2a +1)
(iv)x(x + 4)+ 3x(2x ² – 1)+ 4x ² + 4
(v)a(bc)– b(c – a)– c(a – b)
(vi)a(b – c)+ b(c – a)+ c(a – b)
(vii)4ab(a – b)– 6a ² (b – b ² )-3b ² (2a ² – a)+ 2ab(ba)
(viii)x ² (x ² +1)– x ³ (x +1)– x(x ³ – x)
(ix)2a ² + 3a(1 – 2a ³ )+ a(a + 1)
(x)a ² (2a – 1)+ 3a + a ³ – 8
(xi)a ² b(a – b ² )+ ab ² (4ab – 2a ² )– a ³ b(1 – 2b)
(xii)a ² b(a ³ – a + 1)– ab(a ⁴ – 2a ² + 2a)– b(a ³ – a ² -1)

解决方案：

(i)2x ² (在¹ – x处)– 3x(x ⁴ + 2x)– 2(x ⁴ – 3x ² )

Using Distributive law,

2x² (x³ – x) – 3x (x⁴ + 2x) -2 (x⁴ – 3x²) = 2x⁵ – 2x³ – 3x⁵ – 6x² – 2x⁴ + 6x²

= -x⁵ – 2x⁴ – 2x³

Hence, the product is -x⁵ – 2x⁴ – 2x³

(ii)x ³ y(x ² – 2x)+ 2xy(x ³ – x ⁴ )

Using Distributive law,

x³y (x² – 2x) + 2xy (x³ – x⁴) = x⁵y – 2x⁴y + 2x⁴y – 2x⁵y

= -x⁵y 

Hence, the product is -x⁵y

(iii)3a ² + 2(a + 2)– 3a(2a +1)

Using Distributive law,

3a² + 2(a + 2) – 3a(2a + 1) = 3a² + 2a + 4 – 6a² – 3a

= – 3a² – a + 4

Hence, the product is – 3a² – a + 4

(iv)x(x + 4)+ 3x(2x ² – 1)+ 4x ² + 4

Using Distributive law,

x (x + 4) + 3x (2x² – 1) + 4x² + 4 = x² + 4x + 6x³ -3x + 4x² + 4 

= 6x³ + 5x² + x + 4

Hence, the product is 6x³ + 5x² + x + 4

(v)a(b – c)– b(c – a)– c(a – b)

Using Distributive law,

a (b – c) – b (c – a) – c (a – b) = ab – ac – bc + ab – ac + bc

= 2ab – 2ac

Hence, the product is 2ab – 2ac

(vi)a(b – c)+ b(c – a)+ c(a – b)

Using Distributive law,

a (b – c) + b (c – a) + c (a – b) = ab -ac +bc -ab +ac -bc = 0

Hence, the product is 0

(vii)4ab(a – b)– 6a ² (b – b ² )– 3b ² (2a ² – a)+ 2ab(b – a)

Using Distributive law,

4ab (a – b) – 6a² (b – b²) -3b² (2a² – a) + 2ab (b-a) = 4a²b – 4ab² – 6a²b +6a²b² -6a²b²+3ab² +2ab² -2a²b

= -4a²b + ab²

Hence, the product is -4a²b + ab²

(viii)x ² (x ² +1)– x ³ (x +1)– x(x ³ – x)

Using Distributive law,

x² (x² + 1) – x³ (x + 1) – x (x³ – x) = x⁴ + x² – x⁴ – x³ – x⁴ + x² 

= – x⁴ – x³+ 2x²

Hence, the product is – x⁴ – x³+ 2x²

(ix)2a ² + 3a(1 – 2a ³ )+ a(a + 1)

Using Distributive law,

2a² + 3a (1 – 2a³) + a (a + 1) = 2a² + 3a – 6a⁴ + a²+ a

= – 6a⁴ + 3a² + 4a

Hence, the product is – 6a⁴ + 3a² + 4a

(x)a ² (2a – 1)+ 3a + a ³ – 8

Using Distributive law,

a² (2a – 1) + 3a + a³ – 8 = 2a³ – a² +3a + a³ – 8

= 3a³ – a² + 3a -8

Hence, the product is 3a³ – a² + 3a -8

(xi)a ² b(a – b ² )+ ab ² (4ab – 2a ² )– a ³ b(1 – 2b)

Using Distributive law,

a²b (a – b²) + ab² (4ab – 2a²) – a³b (1 – 2b) = a³b – a²b³ + 4a²b³ – 2a³b² -a³b + 2a³b²

= 3a²b³

Hence, the product is 3a²b³

(xii)a ² b(a ³ – a + 1)– ab(a ⁴ – 2a ² + 2a)– b(a ³ – a ² -1)

Using Distributive law,

a²b (a³ – a + 1) – ab (a⁴ – 2a² + 2a) – b (a³– a² -1) = a⁵b – a³b + a²b – a⁵b + 2a³b – 2a²b – a³b + a²b + b

= b

Hence, the product is b