\(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\)

\(\Leftrightarrow x^3+8y^3=0\)

\(\Leftrightarrow x^3=-8y^3\)

\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\)

\(\Leftrightarrow x^3-8y^3=16\)

\(\Leftrightarrow-8y^3-8y^3=16\)

\(\Leftrightarrow y^3=-1\Rightarrow y=-1\Rightarrow x=2\)

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

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

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

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

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

\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(3-2x\right)\left(4x^2+6x+9\right)\)

\(M=\left(x^3+3^3\right)-\left[3^3-\left(2x\right)^3\right]\)

\(M=x^3+27-27+8x^3\)

\(M=9x^3\)

Thay x=20 vào M ta có:
\(M=9\cdot20^3=72000\)

Vậy: ...

\(N=\left(x-2y\right)\left(x^2+2xy+4y^2\right)+16y^3\)

\(N=x^3-\left(2y\right)^3+16y^3\)

\(N=x^3-8y^3+16y^3\)

\(N=x^3+8y^3\)

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

Thay \(x+2y=0\) vào N ta có:

\(N=0\cdot\left(x^2-2xy+4y^2\right)=0\)

Vậy: ...

từ mik sửa lại mik hơi cấn một tý

\(\left(x-2y\right)^3-\left(x+2y\right)\left(x^2-2xy+4y^2\right)+6x^2y\\ x^3-6x^2+12xy^2-8y^3-\left(x^3+8y^3\right)+6x^2y\\ x^3-6x^2+12xy^2-8y^3-x^3-8y^3+6x^2y\\ =12xy^2-16y^3\)

=x^3-6x^2y+12xy^2-8y^3-x^3-8y^3+6x^2y

=-16y^3+12xy^2

a: \(\dfrac{1}{2}x^2\cdot2x^3-4x^2+3=x^5-4x^2+3\)

b: \(2y\left(xy-1\right)\left(xy+1\right)=2y\left(x^2y^2-1\right)=2x^2y^3-2y\)

a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(=x^3+8y^3-x^3+y^3\)

\(=9y^3\)

b) Ta có: \(\left(x+1\right)\left(x-1\right)^2-\left(x+2\right)\left(x^2-2x+4\right)\)

\(=\left(x+1\right)\left(x^2-2x+1\right)-\left(x+2\right)\left(x^2-2x+4\right)\)

\(=x^3-2x^2+x+x^2-2x+1-\left(x^3+8\right)\)

\(=x^3-x^2-x+1-x^3-8\)

\(=-x^2-x-7\)

\(a,VP=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ =\left(x+2y\right)\left[x^2-x.2y+\left(2y\right)^2\right]\\ =x^3+\left(2y\right)^3=x^3+8y^3=VT\left(đpcm\right)\\ b,VT=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)\\ =x^3-y^3-3xy\left(x-y\right)\\ =x^3-3x^2y+3xy^2-y^3\\ =\left(x-y\right)^3=VP\left(đpcm\right)\)

\(c,VT=\left(x-3y\right)\left(x^2+3xy+9y^2\right)-\left(3y+x\right)\left(9y^2-3xy+x^2\right)\\ =\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]-\left(x+3y\right).\left[x^2-x.3y+\left(3y\right)^2\right]\\ =x^3-27y^3-\left(x^3+27y^3\right)\\ =-54y^3=VP\left(đpcm\right)\)

Chọn A

( x - 2 y ) x 2 + 2 x y + 4 y 2 = ( x ) 3 - ( 2 y ) 3 = x 3 - 8 y 3 .

Ta có :  ( x - 2 y ) ( x 2 + 2 x y + 4 y 2 ) = ( x ) 3 - ( 2 y ) 3 = x 3 - 8 y 3 .

Ta có :  ( x - 2 y ) x 2 + 2 x y + 4 y 2 = ( x ) 3 - ( 2 y ) 3 = x 3 - 8 y 3 .