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

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

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

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

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

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

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

nhìu giữ cha !!!!

a.

$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.

$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.

$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$

d.

$x^3-3x^2+3x-1=(x-1)^3$

e.

$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$

$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$

f.

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

g.

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

h. Không phân tích được thành nhân tử

i.

$x^2-2x-3=(x^2-3x)+(x-3)=x(x-3)+(x-3)=(x+1)(x-3)$

j.

$x^2+4x-12=(x^2-2x)+(6x-12)=x(x-2)+6(x-2)=(x-2)(x+6)$

k.

$x^2-8x-9=(x^2+x)-(9x+9)=x(x+1)-9(x+1)=(x+1)(x-9)$

l.

$x^2+x-6=(x^2+3x)-(2x+6)=x(x+3)-2(x+3)=(x-2)(x+3)$

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

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

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

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

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

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

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

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

\(M=9x^3\)

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: ...

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

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

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

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

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

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

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

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

\(\Rightarrow x^2+4y^2+4xy=x^2y^2+2xy+1-1\)

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

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

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

Vì x,y là các số nguyên nên \(\left(xy-x-2y+1\right),\left(xy+x+2y+1\right)\) là các ước số của 1. Do đó ta có 2 trường hợp:

TH1: \(\left\{{}\begin{matrix}xy-x-2y+1=1\\xy+x+2y+1=1\left(1\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}-xy+x+2y-1=-1\\xy+x+2y+1=1\end{matrix}\right.\)

\(\Rightarrow2\left(x+2y\right)=0\Rightarrow x=-2y\)

Thay vào (1) ta được:

\(-2y^2+1=1\Leftrightarrow y=0\Rightarrow x=0\)

TH2: \(\left\{{}\begin{matrix}xy-x-2y+1=-1\\xy+x+2y+1=-1\left(1\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}-xy+x+2y-1=1\\xy+x+2y+1=-1\end{matrix}\right.\)

\(\Rightarrow2\left(x+2y\right)=0\Rightarrow x=-2y\)

Thay vào (1) ta được:

\(-2y^2+1=-1\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)

\(y=1\Rightarrow x=-2;y=-1\Rightarrow x=2\)

Vậy các cặp số nguyên (x;y) thỏa điều kiện ở đề bài là \(\left(0;0\right),\left(2;-1\right)\left(-2;1\right)\)

c

c

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