\(\left(\dfrac{x}{x+2y}-\dfrac{x+2y}{2y}\right)\left(\dfrac{x}{x-2y}-1+\dfrac{8y^3}{8y^3-x^3}\right)=\dfrac{2xy-\left(x+2y\right)^2}{2y\left(x+2y\right)}\left(\dfrac{2y}{x-2y}+\dfrac{8y^3}{\left(2y-x\right)\left(4y^2+2yx+x^2\right)}\right)=\dfrac{-\left(x^2+2xy+4y^2\right)}{2y\left(x+2y\right)}\cdot\dfrac{2y\left(4y^2+2yx+x^2\right)-8y^3}{\left(x-2y\right)\left(x^2+2xy+4y^2\right)}=\dfrac{-\left(x^2+2xy+4y^2\right)2y\left(4y^2+2xy+x^2-4y^2\right)}{2y\left(x+2y\right)\left(x-2y\right)\left(x^2+2x+4y^2\right)}=\dfrac{-\left(x^2+2xy\right)}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{x}{2y-x}\)

\(\left(x^3+8y^3\right)\div\left(x+2y\right)\)

\(=\left[x^3+\left(2y\right)^3\right]\div\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\div\left(x+2y\right)\)

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

\(\frac{x^3+8y^3}{x+2y}\)

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

\(=\frac{\left(x+2y\right)\left(x^2+4y^2-2xy\right)}{x+2y}\)

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

\(\left(x^3+8y^3\right):\left(x+2y\right)\)

\(=\left[x^3+\left(2y\right)^3\right]:\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x^2-2xy+4y^2\right):\left(x+2y\right)\)

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

a) \(\left(3x+y-z\right)-\left(4x-2y+6z\right)\)

\(=3x+y-z-4x+2y-6z\)

\(=-x+3y-7z\)

b) \(\left(x^3+6x^2+5y^3\right)-\left(2x^3-5x+7y^3\right)\)

\(=x^3+6x^2+5y^3-2x^3+5x-7y^3\)

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

c) \(\left(5,7x^{2y}-3,1xy+8y^3\right)-\left(6,9xy-2,3x^{2y}-8y^3\right)\)

\(=5,7x^{2y}-3,1xy+8y^3-6,9xy+2,3x^{2y}+8y^3\)

\(=8x^{2y}-10xy+16y^3\)

bạn ơi cho mình hỏi phía sau bạn có ghi thiếu gì không

x^3 - 8y^3 - 12xy = (x -2y)(x^2+2xy+4y^2)-12xy

=2x^2+4xy+8y^2-12xy

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

=2(x-2y)^2

the x-2y= 2 vao bieu thuc ta duoc 

x^3-8y^3-12xy=2(2^2)=8

vay gia tri cua bt la 8

P= 125x^3-8y^3

  =5^3x^3-2^3y^3

  =(5x)^3-(2y)^3

  =(5x-2y)(25x^2+10xy+4y^2)

P=4x(x-2y)+8y(2y-x)

  =4x(x-2y)-8y(x-2y)

  =(4x-8y)(x-2y)

  =4(x-2y)(x-2y)

  =4(x-2y)^2

(2x+1)^2-(x-1)^2=(2x+1-x+1)(2x+1+x-1)

                       =(x+2)3x

K NHA!