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Given,

a + b = 10, ab = 21 

Choose a + b = 10 

Cubing both sides, 

(a + b)³ = (10)³ 

a³ + b³ + 3ab(a + b) = 1000 

a³ + b³ + 3 x 21 x 10 = 1000 

a³ + b³ + 630 = 1000 

a³ + b³ = 1000 – 630

a³ + b³ = 370

(i) 51² − 49² = (51 + 49) (51 − 49)

= (100) (2) = 200

(ii) (1.02)² − (0.98)² = (1.02 + 0.98) (1.02 − 0.98) = (2) (0.04) = 0.08

Given ,

x – y

We know that

x² + y² = 29

x² + y² + 2xy – 2xy = 29

(x – y)² + 2 (2) = 29

(x – y)² + 4 = 29

(x – y)² = 25

(x – y) = ± 5

(i) (x + 4) (x + 7)

= x (x + 7) + 4 (x + 7)

= x² + 7x + 4x + 28

= x² + 11x + 28

(ii) (x – 11) (x + 4

= x (x + 4) – 11 (x + 4)

= x² + 4x – 11x – 44

= x² – 7x – 44

(iii) (x + 7) (x – 5)

= x (x – 5) + 7 (x – 5)

= x² – 5x + 7x – 35

= x² + 2x – 35

(i) (x – 3) (x – 2)

= x (x – 2) – 3 (x – 2)

= x² – 2x – 3x + 6

= x² – 5x + 6

(ii) (y² – 4) (y² – 3)

= y² (y² – 3) – 4 (y² – 3)

= y⁴ – 3y² – 4y² + 12

= y⁴ – 7y² + 12

(iii) (x + 4/3) (x + 3/4)

= x (x + 3/4) + 4/3 (x + 3/4)

= x² + 3x/4 + 4x/3 + 12/12

= x² + 3x/4 + 4x/3 + 1

= x² + 25x/12 + 1

Let us simplify the given expression

= 2.3a⁵b² × 1.2a²b²

= 2.3 × 1.2 × a⁵ × a² × b² × b²

= 2.76 × a⁵⁺² × b²⁺²

= 2.76a⁷b⁴

Now let us substitute when, a = 1 and b = 0.5

For 2.76 a7 b4

= 2.76 (1)⁷ (0.5)⁴

= 2.76 × 1 × 0.0025

= 0.1725

= 6.9/40

(i) (3x² – 4xy) (3x² – 3xy)

= 3x² (3x² – 3xy) – 4xy (3x² – 3xy)

= 9x⁴ – 9x³y – 12x³y + 12x²y²

= 9x⁴ – 21x³y + 12x²y²

(ii) (x + 1/5) (x + 5)

= x (x + 1/5) + 5 (x + 1/5)

= x² + x/5 + 5x + 1

= x² + 26/5x + 1

(iii) (z + 3/4) (z + 4/3)

= z (z + 4/3) + 3/4 (z + 4/3)

= z² + 4/3z + 3/4z + 12/12

= z² + 4/3z + 3/4z + 1

= z² + 25/12z + 1

(iv) (x² + 4) (x² + 9)

= x² (x² + 9) + 4 (x² + 9)

= x⁴ + 9x² + 4x² + 36

= x⁴ + 13x² + 36

(v) (y² + 12) (y² + 6)

= y² (y² + 6) + 12 (y² + 6)

= y⁴ + 6y² + 12y² + 72

= y⁴ + 18y² + 72

(vi) (y² + 5/7) (y² – 14/5)

= y² (y² – 14/5) + 5/7 (y² – 14/5)

= y⁴ – 14/5y² + 5/7y² – 2

= y⁴ – 73/35y² – 2

(i) (3x + 5) (3x + 11)

= 3x (3x + 11) + 5 (3x + 11)

= 9x² + 33x + 15x + 55

= 9x² + 48x + 55

(ii) (2x² – 3) (2x² + 5)

= 2x² (2x² + 5) – 3 (2x² + 5)

= 4x⁴ + 10x² – 6x² – 15

= 4x⁴ + 4x² – 15

(iii) (z² + 2) (z² – 3)

= z² (z² – 3) + 2 (z² – 3)

= z⁴ – 3z² + 2z² – 6

= z⁴ – z² – 6

(iv) (3x – 4y) (2x – 4y)

= 3x (2x – 4y) – 4y (2x – 4y)

= 6x² – 12xy – 8xy + 16y²

= 6x² – 20xy + 16y²

 a²b (a³– a + 1) – ab(a⁴ – 2a² + 2a) – b(a³– a² – 1)

= a⁵b – a³b + a²b – a⁵b + 2a³b – 2a²b – ba³ + a²b + b

= a⁵b – a⁵b – a³b + 2a³b – ba³ + a²b – 2a²b + a²b + b

= b

(i)  -8/27xyz (3/2xyz² – 9/4xy²z³)

= -8/27xyz (3/2xyz² – 9/4xy²z³)

= (-8/27xyz × 3/2xyz²) – (-8/27xyz × 9/4xy²z³)

= -4/9x²y²z³ + 2/3x²y³z⁴

(ii) -4/27xyz (9/2x²yz – 3/4xyz²)

= -4/27xyz (9/2x²yz – 3/4xyz²)

= (-4/27xyz × 9/2x²yz) – (-4/27xyz × 3/4xyz²)

= -2/3x³y²z² + 1/9x²y²z³

(iii) 1.5x (10x2y – 100xy2)

= 1.5x (10x²y – 100xy²)

= (1.5x 10x²y) – (1.5x × 100xy²)

= 15x³y – 150x²y²

(i)  6x/5(x³ + y³)

= 6/5x (x³ + y³)

= (6/5x × x³) + (6/5x × y³)

= 6/5x⁴ + 6/5xy³

(ii) xy (x³ – y³)

= xy (x³ – y³)

= (xy × x³) – (xy × y³)

= x⁴y – xy⁴

(iii) 0.1y (0.1x⁵ + 0.1y)

= 0.1y (0.1x5 + 0.1y)

= (0.1y × 0.1x⁵) + (0.1y × 0.1y)

= 0.01x⁵y + 0.01y²

(iv) (-7/4ab²c – 6/25a²c²) (-50a²b²c²)

= (-7/4ab²c – 6/25a²c²) (-50a²b²c²)

= (-7/4ab²c × -50a²b²c²) – (6/25a²c² × -50a²b² × c²)

= 350/4a³b⁴c³ + 12a⁴b²c⁴

= 175/2a³b⁴c³ + 12a⁴b²c⁴

(i) We know, (a - b)² = a² + b² -2ab

Here,

a = 2x and b = \(\frac{1}{x}\)

\((2x-\frac{1}{x})^2\) = \((2x)^2-(\frac{1}{x})^2 - 2\times x\times \frac{1}{x}\)

= \((4x)^2-(\frac{1}{x})^2 - 2\)

(ii) We know, (a-b)² = (a+b) (a - b)

(2x+y)(2x-y) = (2x)²– y² = 4x² – y²

(iii) We know, (a - b)² = a² + b² - 2ab

Here,

(a²b - b²a)² = (a²b)² + (b²a)² - 2 x a²b x b²a

= a⁴b² + a²b⁴ - 2a³b³

(iv) We know, (a-b)² = (a+b) (a - b)

(a – 0.1)(a +0.1) = (a)²– (0.1)²

= a² – 0.01

(v) We know, (a - b)² = a² - b²

(1.5x² - 0.3y²)(1.5x² - 0.3y²) = (1.5x²)² - (0.3y²)²

= 2.25x⁴ - 0.09y⁴

2x² – 3y², 5x² + 6y², -3x² – 4y²

Required sum,

= (2x² – 3y²) + (5x² + 6y²) + (-3x² – 4y²)

Collecting like terms,

= 2x² + 5x² – 3x² – 3y² + 6y² – 4y²

Adding like terms,

= (2 + 5 – 3)x² + (– 3 + 6 – 4)y²

= 4x² – y²

Given 5x-8y+2z, 3z-4y-2x, 6y-z-x and 3x-2z-3y

To add the given expression we have arrange them column wise is given below:

5x-8y+2z

-2x-4y+3z

-x+6y-z

(i) 5x² × 4x3

= 20 × x⁵

= 20x⁵

(ii)  -3a² × 4b⁴

= – 3 × a² × 4 × b⁴

= -12a²b⁴

(iii) (-5xy) × (-3x2yz)

= 15 × x¹⁺² × y¹⁺¹ × z

= 15x³y²z

(iv) 1/2xy × 2/3x²yz²

= 1/2 × 2/3 × x × x² × y × y × z²

= 1/3 × x¹⁺² × y¹⁺¹ × z²

= 1/3x³y²z²

Now,

 let us simplify the given expression

= (x²y – 1) × (3 – 2x²y)

= x²y (3 – 2x²y) – 1 (3 – 2x²y)

= 3x²y – 2x⁴y² – 3 + 2x²y

= 5x²y – 2x⁴y² – 3

Let us substitute the given values x = – 1 and y = – 2, then

= (x²y – 1) × (3 – 2x²y)

= [(-1)² (-2) – 1] × [3 – 2 (-1)² (-2)

= (-2 – 1) × (3 + 4)

= -3 × 7

= -21

Now,

= 5x²y – 2x⁴y² – 3

= [-2 (-1)⁴ (-2)² + 5 (-1)² (2) – 3]

= – 8 – 10 – 3

= -21

∴ The given expression is verified.

(i) (1/3y² – 4/7y + 11) – (1/7y – 3 + 2y²) – (2/7y – 2/3y² + 2)

= 1/3y² – 2y² – 2/3y² – 4/7y – 1/7y – 2/7y + 11 + 3 – 2

= (y² – 6y² + 2y²)/3 – (4y – y – 2y)/7 + 12

= -3/3y² – 7/7y + 12

= -y² – y + 12

(ii) -1/2a²b²c + 1/3ab²c – 1/4abc² – 1/5cb²a² + 1/6cb²a – 1/7c²ab + 1/8ca²b

= -1/2a²b2c – 1/5a²b²c + 1/3ab²c + 1/6ab²c – 1/4abc² – 1/7abc² + 1/8a²bc

= -7/10a²b²c + 1/2ab²c – 11/28abc² + 1/8a²bc

 (i) x² (x – y) y² (x + 2y)

= x² (x – y) y² (x + 2y)

= x²y² (x² + 2xy – xy – 2y²)

= x²y² (x² + xy – 2y²)

= x⁴y² + x³y³ – 2x²y⁴

(ii)  (x³ – 2x² + 5x – 7) (2x – 3)

= (x³ – 2x² + 5x – 7) (2x – 3)

= 2x⁴ – 4x³ + 10x² – 14x – 3x³ + 6x² – 15x + 21

= 2x⁴ – 7x³ + 16x² – 29x + 21

(iii) (5x + 3) (x – 1) (3x – 2)

= (5x + 3) (x – 1) (3x – 2)

= (5x² – 2x – 3) (3x – 2)

= 15x³ – 6x² – 9x – 10x² + 4x + 6

= 15x³ – 16x² – 5x + 6

Given,

 x² (x + 2y) (x – 3y)

= x² (x² – 3xy + 2xy – 3y²)

= x² (x² – xy – 6y²)

= x⁴ – x³y – 6x²y²

Now,

 let us simplify the given expression

= (x² – 2y²) (x + 4y)x²y²

= (x³ + 4x²y – 2xy² – 8y³) × x²y²

= x⁵y² + 4x⁴y³ – 2x³y⁴ – 8x²y⁵

Now,

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

= x⁴ – 2x³ + 3x² – 4x – x³ + 2x² – 3x + 4 – (2x³ – 2x² + 2x – 3x² + 3x – 3)

= x⁴ – 3x³ + 5x² – 7x + 4 – 2x³ + 5x² – 5x + 3

= x⁴ – 5x³ + 10x² – 12x + 7

(i) (5 – x) (6 – 5x) (2 – x)

= (5 – x) (6 – 5x) (2 – x)

= (x² – 7x + 10) (6 – 5x)

= -5x³ + 35x² – 50x + 6x² – 42x + 60

= 60 – 92x + 41x² – 5x³

(ii) (2x² + 3x – 5) (3x² – 5x + 4)

= (2x² + 3x – 5) (3x² – 5x + 4)

= 6x⁴ + 9x³ – 15x² – 10x³ – 15x² + 25x + 8x² + 12x – 20

= 6x⁴ – x³ – 22x² + 37x – 20

(iii) (3x – 2) (2x – 3) + (5x – 3) (x + 1)

= (3x – 2) (2x – 3) + (5x – 3) (x + 1)

= 6x² – 9x – 4x + 6 + 5x² + 5x – 3x – 3

= 11x² – 11x + 3

(i) x² – 3x + 5 – 1/2(3x² – 5x + 7)

= x² – 3/2x² – 3x + 5/2x + 5 – 7/2

= (2x² – 3x²)/2 – (6x + 5x)/2 + (10-7)/2

= -1/2x² – 1/2x + 3/2

(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)

= 5 – 3x + 2y – 2x + y – 3x + 7y – 9

= – 3x – 2x – 3x + 2y + y + 7y + 5 – 9

= -8x + 10y – 4

(iii) 11/2x²y – 9/4xy² + 1/4xy – 1/14y²x + 1/15yx² + 1/2xy

= 11/2x²y + 1/15x²y – 9/4xy² – 1/14xy² + 1/4xy + 1/2xy

= (165x²y + 2x²y)/30 + (-126xy² – 4xy²)/56 + (xy + 2xy)/4

= 167/30x²y – 130/56xy² + 3/4xy

= 167/30x²y – 65/28xy² + 3/4xy

(i) x(x+4) + 3x(2x² -1) + 4x² + 4

= x² + 4x + 6x³ – 3x + 4x² + 4

= 6x³ + 5x² + x + 4

(ii) a(b-c) – b(c-a) – c(a-b)

= ab – ac – bc + ab – ca + bc

= 2ab – 2ac

(iii) a(b-c) +b(c-a) + c(a-b)

= ab – ac + bc – ab + ac – bc

= 0

Now,

= (3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)

= 12x² + 9xy + 8xy

= 12x² + 9xy + 8xy + 6y² – 14x² + 6xy + 7xy – 3y²

= -2x² + 3y² + 30xy

Now,

let us simplify the given expression

= (x² – 3x + 2) (5x – 2) – (3x² + 4x – 5) (2x – 1)

= 5x³ – 15x² + 10x – 2x² + 6x – 4 – (6x³ + 8x² – 10x – 3x² – 4x + 5)

= 5x³ – 6x³ – 15x² – 2x² – 5x² + 16x + 14x – 4 – 5

= – x³ – 22x² + 30x – 9

Now,

= (5x – 3) (x + 2) – (2x + 5) (4x – 3)

= 5x² + 10x – 3x – 6 – 8x² + 6x – 20x + 15

= -3x2 – 7x + 9

Given,

9.8 × 10.2

We can express 9.8 as 10 – 0.2 and 10.2 as 10 + 0.2

Using formula (a – b) (a + b) = a² – b²

We get,

9.8 × 10.2

= (10 – 0.2) (10 + 0.2)

= (10)² – (0.2)²

= 100 – 0.04

= 99.96

(i) ((58)² – (42)²)/16

Using formula (a – b) (a + b) = a² – b²

We get,

((58)² – (42)²)/16 

= ((58-42) (58+42)/16)

= ((16) (100)/16)

= 100

(ii) 178 × 178 – 22 × 22

Using formula (a – b) (a + b) = a² – b²

We get,

178 × 178 – 22 × 22 

= (178)² – (22)²

= (178-22) (178+22)

= 200 × 156

= 31200

(i) (1001)²

 Hence we take as , 

1001 as 1000 + 1

So, 

= (1001)² 

= (1000 + 1)²

= (1000)² + 2 (1000) (1) + 1²

= 1000000 + 2000 + 1

= 1002001

(ii) (999)²

Hence we take as ,

999 as 1000 – 1

So, 

= (999)² 

= (1000 – 1)²

= (1000)² – 2 (1000) (1) + 1²

= 1000000 – 2000 + 1

= 998001

(i) 7x²yz – 5xy

The given equation has two terms that are:

7x²yz and – 5xy

The coefficient of 7x²yz is 7

The coefficient of – 5xy is – 5

(ii) x² + x + 1

The given equation has three terms that are:

x², x, 1

The coefficient of x² is 1

The coefficient of x is 1

The coefficient of 1 is 1

(iii) 3x²y² – 5x²y²z² + z²

The given equation has three terms that are:

3x²y, -5x²y²z² and z²

The coefficient of 3x²y is 3

The coefficient of -5x²y²z² is -5

The coefficient of z² is 1

(iv) 9 – ab + bc – ca

The given equation has four terms that are:

9, -ab, bc, -ca

The coefficient of 9 is 9

The coefficient of -ab is -1

The coefficient of bc is 1

The coefficient of -ca is -1

(v) a/2 + b/2 – ab

The given equation has three terms that are:

a/2, b/2, -ab

The coefficient of a/2 is 1/2

The coefficient of b/2 is 1/2

The coefficient of -ab is -1

(vi) 0.2x – 0.3xy + 0.5y

The given equation has three terms that are:

0.2x, -0.3xy, 0.5y

The coefficient of 0.2x is 0.2

The coefficient of -0.3xy is -0.3

The coefficient of 0.5y is 0.5

Let us we solve

= -8 × -20 × x² × x × y⁶ × y

= 160 × x²⁺¹ × y⁶⁺¹

= 160x³y⁷

Now,

 let us verify when, x = 2.5 and y = 1

For 160x³y⁷

= 160 (2.5)³ × (1)⁷

= 16 × 15.625

= 250

For -8x²y⁶ and -20xy

= -8 × 2.52 × 16 × -20 × 1 × 2.5

= 250

Hence, the given expression is verified.

Given ,

(x²)³ × (2x) × (-4x) × (5)

= 2 × -4 × 5 × x⁶ × x × x

= -40 × x⁶⁺¹⁺¹

= -40x⁸

(i) 4y – 7z

This expression is binomial because it contains two terms. That is 4 y and -7z.

(ii) y²

This is monomial expression because it contains only one term y².

(iii) x + y – xy

This is trinomial expression because it contains three terms x, y and -xy.

(iv) 100

This is a monomial expression because it contains only one term 100.

(v) ab – a – b

This is trinomial expression because it contains 3 terms ab, -a and -b.

(vi) 5 – 3t

This is binomial expression because it contains 2 terms 5 and -3t.

(vii) 4p²q – 4pq²

This is binomial expression because it contains two terms 4p²q and – 4 pq²

(viii) 7mn

This is monomial expression because it contains only one term 7mn.

(ix) z² – 3z + 8

This is trinomial expression because it contains three terms z², -3z and 8.

(x) a² + b²

This is binomial expression because it contains two terms a² and b².

(xi) z² + z

This is binomial expression because it contains 2 terms z² and z.

(xii) 1 + x + x²

This is trinomial expression because it contains 3 terms 1, x and x².

(i) (3x) × (4x) × (-5x)

= -60 × x¹⁺¹⁺¹

= -60x³

(ii) (4x²) × (-3x) × (4/5x³)

= -48/5 × x²⁺¹⁺³

= -485x⁶

(iii) (5x4) × (x²)³ × (2x)²

= 5 × 4 × x⁴ × x⁶ × x²

= 20 × x⁴⁺⁶⁺²

= 20x¹²

(i) (1/8x²y⁴) × (1/4x⁴y²) × (xy) × 5

= 1/8 × 1/4 × 5 × x² × x⁴ × x × y⁴ × y² × y

= 5/32 × x²⁺⁴⁺¹ × y⁴⁺²⁺¹

= 5/32x⁷y⁷

Now let us substitute when, x = 1 and y = 2

= 5/32 × 1⁶ × 2⁶

= 5/32 × 64

= 5 × 2

= 10

(ii) (2/5a²b) × (-15b²ac) × (-1/2c²)

= 2/5 × -15 × -1/2 × a² × a × b × b² × c × c²

= 3 × a²⁺¹ × b¹⁺² × c¹⁺²

= 3a³b³c³

(iii) (-4/7a²b) × (-2/3b²c) × (-7/6c²a)

= -4/7 × -2/3 × -7/6 × a² × a × b × b² × c × c²

= -4/9 × a²⁺¹ × b²⁺¹ × c¹⁺²

= -4/9a³b³c³

(i) -5xy from 12xy

= 12xy – (- 5xy)

= 5xy + 12xy

= 17xy

(ii) 2a2 from -7a2

= 2a² + (-7a²)

= -2a² + 7a²

= -9a²

(iii) 2a - b from 3a-5b

= -(2a – b)+ (3a – 5b)

= -2a + b+ 3a – 5b

= a – 4b

(iv) 2x³ – 4x² + 3x + 5 from 4x³ + x² + x + 6

= – (2x³ – 4x² + 3x + 5) + (4x³ + x² + x + 6)

= – 2x³ + 4x² – 3x – 5 + 4x³ + x² + x + 6

= 2x³ + 5x² – 2x + 1

(v) 3/2y³ – 2/7y² – 5 from 1/3y³ + 5/7y² + y – 2

= 1/3y³ + 5/7y² + y – 2 – 2/3y³ + 2/7y² + 5

= 1/3y³ – 2/3y³ + 5/7y² + 2/7y² + y – 2 + 5

= -1/3y³ + 7/7y² + y + 3

= -1/3y³ + y² + y + 3

(i) (7/9ab²) × (15/7ac2b) × (-3/5a²c)

= 7/9 × 15/7 × -3/5 × a × a × a² × b² × b × c² × c

= -a⁴b³c³

(ii) (4/3u²vw) × (-5uvw²) × (1/3v²wu)

= -20/9 × u²⁺¹⁺¹ × v¹⁺¹⁺² × w¹⁺²⁺¹

= -20/9u⁴v⁴w⁴

(iii) (0.5x) × (1/3xy²z⁴) × (24x²yz)

= 12/3 × x¹⁺¹⁺² × y²⁺¹ × z⁴⁺¹

= 4x⁴ × y³ × z⁵

= 4x⁴y³z⁵

(i) 15y² (2 – 3x)

= 30y² – 45xy²

By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000

= 30 × (25/100)² – 45 × (-1) × (25/100)²

= 30 (1/16) + 45 (1/16)

= 15/8 + 45/16

= (30+45)/16

= 75/16

(ii) -3x (y² + z²)

= -3xy² + -3xz²

By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000

= -3× (-1) × (25/100)² – 3 × (-1) × (5/1000)²

= (3×25×25/100×100) + (3×5×5/1000×1000)

= 3/16 + 3/40000

= 39/200

let us simplify the given expression

= (2x² – 1) × (4x³ + 5x²)

= 2x² (4x³ + 5x²) – 1 (4x³ + 5x²)

= 8x⁵ + 10x⁴ – 4x³ – 5x²

Given ,

64x² + 81y² + 144xy when x = 11 and y = 4/3

Using formula (a + b)² = a² + b² +2ab

(8x)² + 2 (8x) (9y) + (9y)² (8x + 9y)

= [8 (11) + 9 (4/3)]²

= (88 + 12)²

= (100)²

= 10000

Let us simplify the given expression

x³z + xzy²

By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000

= x³z + xzy²

= (-1)³ × (5/1000) + (-1) × (5/1000) × (25/100)²

= -1/200 – 1/16 × 1/200

= -1/200 – 1/3200

= (-16 -1)/3200

= -17/3200