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12 tháng 4 2021

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

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

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

AH
Akai Haruma
Giáo viên
23 tháng 8 2019

Lời giải:

Những bài này sử dụng những hằng đẳng thức đáng nhớ.

Vì $x=-2$ nên $x+2=0$. Ta có:

\(A=(2x-3)^2-(x-3)^3+(4x+1)[(4x)^2-4x.1+1^2]\)

\(=(2x-3)^2-(x-3)^3+(4x)^3+1^3\)

\(=[2(x+2)-7]^2-(x+2-5)^3+8x^3+1\)

\(=(-7)^2-(-5)^3+8.(-2)^3+1=111\)

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\(B=(3x-y)^3-[x^3+(2y)^3]+(x+3)^2\)

\(=(3.1-2)^3-(1^3+8.2^3)+(1+3)^2=-48\)

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Vì $x=\frac{1}{2}; y=\frac{-1}{2}\Rightarrow x+y=0$

\(C=(x-5y)^2+(2x-3y)^3-(x-y)^3-[(2x)^3+(3y)^3]\)

\(=(x+y-6y)^2+[2(x+y)-5y]^3-(x+y-2y)^3-[8(x^3+y^3)+19y^3]\)

\(=(-6y)^2+(-5y)^3-(-2y)^3-19y^3\)

\(=36y^2-136y^3=36.(\frac{-1}{2})^2-136(\frac{-1}{2})^3=26\)

30 tháng 10 2021

\(2x^3y-2xy^3-4xy^2-2xy\)

\(=2xy.\left(x^2-y^2-2y-1\right)\)

\(=2xy.[x^2-\left(y^2+2y+1\right)]\)

\(=2xy.[x^2-\left(y+1\right)^2]\)

\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)

Vậy chọn đáp án A

12 tháng 1 2022

chọn A

29 tháng 6 2015

\(\left(x+3y\right)^3-\left(x+3y\right)\left(x^2-3xy+9y^2\right)-2x\left(x-2\right)^2=\left(x+3y\right)^3-\left(x^3+27y^3\right)-2x\left(x-2\right)^2\)

Thay x=1 y=2 ta có:

\(\left(1+3.2\right)^3-\left(1^3+27.2^3\right)-2.1.\left(1-2\right)^2=7^3-\left(1+216\right)-2=343-217-2=124\)

 

30 tháng 11 2019

1) ĐKXĐ: x \(\ne\)1; x \(\ne\)0

Ta có: A = \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x}{x\left(x-1\right)}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{12}{x^2+x+1}\)

b) Ta có: B = \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)

B = \(\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)

B = \(\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x+3y\right)\left(x-3y\right)}\)

B = \(\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B =  \(\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B = \(\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B = \(\frac{x+3y}{x\left(x-3y\right)}\)

30 tháng 11 2019

\(A=\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x\left(1-x\right)}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}-\frac{6x}{x\left(x-1\right)}\)

\(A=\frac{x\left(4x^2-3x+17\right)+x\left(x-1\right)\left(2x-1\right)-6x\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{4x^3-3x^2+17x+x\left(2x^2-x-2x+1\right)-6x^3-6x^2-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{\left(4x^3+2x^3-6x^3\right)-3x^2-3x^3-6x^2+17x+x-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x^2+12x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}\)

12 tháng 7 2017

1) (x+3)(x2- 3x + 9) = x3 + 27
2) (x2 + 2y)2 = x4 + 4xy + 4y2 
3) (2x-3)(2x+3) = 4x2 - 9
4) (x + 3y)3 = x3 + 9x2y + 9xy2 + y3
5) (2x2- y)3 = 8x6 - 6x4y + 6x2y2 - y3
6) (x-3y)(x2 + 3xy +9y2)= x3- 27y3
7) (2x + 3y)(4x2 - 6xy +9y2)= 8x3 + 27y3
8) (3x - y2)2= 9x2 - 6xy2 + y4