Trả lời:

Bài 4:

b, B =  ( x + 1 ) ( x⁷ - x⁶ + x⁵ - x⁴ + x³ - x² + x - 1 ) 

= x⁸ - x⁷ + x⁶ - x⁵ + x⁴ - x³ + x² - x + x⁷ - x⁶ + x⁵ - x⁴ + x³ - x² + x - 1 

= x⁸ - 1

Thay x = 2 vào biểu thức B, ta có:

2⁸ - 1 = 255

c, C = ( x + 1 ) ( x⁶ - x⁵ + x⁴ - x³ + x² - x + 1 ) 

= x⁷ - x⁶ + x⁵ - x⁴ + x³ - x² + x + x⁶ - x⁵ + x⁴ - x³ + x² - x + 1

= x⁷ + 1

Thay x = 2 vào biểu thức C, ta có:

2⁷ + 1 = 129

d, D = 2x ( 10x² - 5x - 2 ) - 5x ( 4x² - 2x - 1 ) 

= 20x³ - 10x² - 4x - 20x³ + 10x² + 5x

= x

Thay x = - 5 vào biểu thức D, ta có:

D = - 5

Bài 5: 

a, A = ( x³ - x²y + xy² - y³ ) ( x + y )

= x⁴ + x³y - x³y - x²y² + x²y² + xy³ - xy³ - y⁴

= x⁴ - y⁴

Thay x = 2; y = - 1/2 vào biểu thức A, ta có:

A = 2⁴ - ( - 1/2 )⁴ = 16 - 1/16 = 255/16

b, B = ( a - b ) ( a⁴ + a³b + a²b² + ab³ + b⁴ ) 

= a⁵ + a⁴b + a³b² + a²b³ + ab⁴ - ab⁴ - a³b² - a²b³ - ab⁴ - b⁵ 

= a⁵ + a⁴b - ab⁴ - b⁵

Thay a = 3; b = - 2 vào biểu thức B, ta có:

B = 3⁵ + 3⁴.( - 2 ) - 3.( - 2 )⁴ - ( - 2 )⁵ = 243 - 162 - 48 + 32 = 65

c, ( x² - 2xy + 2y² ) ( x² + y² ) + 2x³y - 3x²y² + 2xy³ 

= x⁴ + x²y² - 2x³y - 2xy³ + 2x²y² + 2y⁴ + 2x³y - 3x²y² + 2xy³

= x⁴ + 2y⁴

Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:

( - 1/2 )⁴ + 2.( - 1/2 )⁴ = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16

a: \(=15x^4-12x^3+9x^2\)

c: \(=5x^3-15x^2-4x^2+12x\)

\(=5x^3-19x^2+12x\)

\(x^4.\frac{2x^2-x-1}{3}\)

đúng k bn

đúng rồi bạn Thanh Ngân ơi

Giups mình nhé

\(a)-2y(5xy+3x^2)\\=-10xy^2-6x^2y\\b)(2x+y)(4x-y)\\=8x^2-2xy+4xy-y^2\\=8x^2+2xy-y^2\\c)(x-1^2)-(x+1^2)\\=(x-1)-(x+1)\\=x-1-x-1\\=-2(bạn.xem.lại.đề.phần.này)\\d)(xem.lại.đề)\)

a: \(-2y\left(5xy+3x^2\right)=-10xy^2-6x^2y\)

b: \(\left(2x+y\right)\left(4x-y\right)\)

\(=8x^2-2xy+4xy-y^2\)

\(=8x^2+2xy-y^2\)

`a)A=x(x+y)-x(y-x)`

`=x^2+xy-xy+x^2`

`=2x^2`

Thay `x=-3`

`=>A=2.9=18`

`b)B=4x(2x+y)+2y(2x+y)-y(y+2x)`

`=8x^2+4xy+4xy+2y^2-y^2-2xy`

`=8x^2+y^2+6xy`

Thay `x=1/2,y=-3/4`

`=>B=8*1/4+9/16-9/4`

`=2+9/16-9/4`

`=9/16-1/4=5/16`

Giúp với mọi người

c: \(=3x^3-4x^2+6x^2-8x=3x^2+2x^2-8x\)

ĐKXĐ: \(a\ne1\)

a. \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)

\(=\dfrac{3a^2-a+3+\left(1-a\right).\left(a-1\right)-2.\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{3a^2-a+3-a^2+2a-1-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{-a+1}{\left(a-1\right).\left(a^2+a+1\right)}\)

\(=-\dfrac{1}{a^2+a+1}\)

a) Ta có: \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)

\(=\dfrac{3a^2-a+3}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{\left(a-1\right)^2}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{2\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{3a^2-a+3-\left(a^2-2a+1\right)-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{a^2-3a+1-a^2+2a-1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{-a}{\left(a-1\right)\left(a^2+a+1\right)}\)

b) Ta có: \(x-\dfrac{xy}{x+y}-\dfrac{x^3}{x^2y^2}\)

\(=x-\dfrac{xy}{x+y}-\dfrac{x}{y^2}\)

\(=\dfrac{xy^2\cdot\left(x+y\right)}{y^2\cdot\left(x+y\right)}+\dfrac{y^2\cdot xy}{y^2\cdot\left(x+y\right)}-\dfrac{x\cdot\left(x+y\right)}{y^2\cdot\left(x+y\right)}\)

\(=\dfrac{x^2y^2+xy^3+xy^3-x^2-xy}{y^2\cdot\left(x+y\right)}\)

\(=\dfrac{x^2y^2+2xy^3-x^2-xy}{y^2\cdot\left(x+y\right)}\)

a: =-4xyz^2

b: =-9x^2y

c: =16x^2y^2

d: =1/6x^2y^3

e: =13/6x^3y^2

f: =7/12x^4y

a) -xyz² - 3xz.yz

= -xyz² - 3xyz²

= -4xyz²

b) -8x²y - x.(xy)

= -8x²y - x²y

= -9x²y

c) 4xy².x - (-12x²y²)

= 4x²y² + 12x²y²

= 16x²y²

d) 1/2 x²y³ - 1/3 x²y.y²

= 1/2 x²y³ - 1/3 x²y³

= 1/6 x²y³

e) 3xy(x²y) - 5/6 x³y²

= 3x³y² - 5/6 x³y²

= 13/6 x³y²

f) 3/4 x⁴y - 1/6 xy.x³

= 3/4 x⁴y - 1/6 x⁴y

= 7/12 x⁴y