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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

(x3+8y3) = (x+2y)(x2- 2xy+4y2)
=> (x3+8y3) : (x+2y) = x2-2xy+4y2

\(\left(x^3+8y^3\right):\left(x+2y\right)=\left(x^3+2^3y^3\right):\left(x+2y\right)=\left[x^3+\left(2y\right)^3\right]:\left(x+2y\right)\)
Áp dụng hằng đẳng thức : \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\left[x^3+\left(2y\right)^3\right]:\left(x+2y\right)=\left[\left(x+2y\right).\left(x^2-2xy+4y^2\right)\right]:\left(x+2y\right)=x^2-2xy+4y^2\)

\(\frac{x^3+8y^3}{x+2y}=\frac{x^3+\left(2y\right)^3}{x+2y}=\frac{\left(x+2y\right)\left(x^2-2xy+4y^2\right)}{x+2y}=x^2-2xy+4y^2\)

a) \(\left(x^3+8y^3\right):\left(x+2y\right)\\ =\left(x+2y\right)\left(x^2-2xy+4y^2\right):\left(x+2y\right)=x^2-2xy+4y^2\)


Ta co: a = x^3 - 8y^3 => a = ( x - 2y ) ( x^2 + 2xy + 4y^2 ) => a = 5. ( 29 + 2xy) ( vi x - 2y = 5 va x^2 + 4y^2 = 29 ) (1)
Mat khac : x - 2y = 59(gt) => ( x - 2y )^2 = 25 => x^2 - 4xy + 4y^2 = 25 => 29 - 4xy = 25 ( vi x^2 + 4y^2 = 29 )
=> xy = 1 (2)

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