\(x^3-y^3-z^3=3xyz\)

=>\(\left(x-y\right)^3-z^3+3xy\left(x-y\right)-3xyz=0\)

=>\(\left(x-y-z\right)\left[\left(x-y\right)^2+z\left(x-y\right)+z^2\right]+3xy\left(x-y-z\right)=0\)

=>\(\left(x-y-z\right)\left[x^2-2xy+y^2+xz-zy+z^2+3xy\right]=0\)

=>\(\left(x-y-z\right)\left(x^2+y^2+z^2+xy+xz-yz\right)=0\)

=>\(\left(x-y-z\right)\left(2x^2+2y^2+2z^2+2xy+2xz-2yz\right)=0\)

=>\(\left(x-y-z\right)\left[\left(x^2+2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2+2xz+z^2\right)\right]=0\)

=>\(\left(x-y-z\right)\left[\left(x+y\right)^2+\left(y-z\right)^2+\left(x+z\right)^2\right]=0\)

=>\(\left[{}\begin{matrix}x-y-z=0\\\left(x+y\right)^2+\left(y-z\right)^2+\left(x+z\right)^2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=y+z\\y=z=-x\end{matrix}\right.\)

\(H=\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1-\dfrac{z}{x}\right)\)

\(=\dfrac{y-x}{y}\cdot\dfrac{z+y}{z}\cdot\dfrac{x-z}{x}\)

TH1: x=y+z

=>\(H=\dfrac{y-x}{y}\cdot\dfrac{x}{z}\cdot\dfrac{x-z}{x}\)

\(=\dfrac{y}{x}\cdot\dfrac{z}{x}\cdot\dfrac{y-x}{y}=\dfrac{y}{x}\cdot\dfrac{z}{x}\cdot\dfrac{-z}{y}=-1\)

TH2: y=z=-x

=>y+x+z=0(vô lý vì x,y,z đều dương)

Vậy: H=-1

Bài 3:

a: \(\left(y-2\right)^3-\left(y-3\right)\left(y^2+3y+9\right)+6\left(y+1\right)^2=49\)

=>\(y^3-6y^2+12y-8-\left(y^3-27\right)+6\left(y^2+2y+1\right)=49\)

=>\(y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)

=>\(24y+25=49\)

=>24y=24

=>y=1

Bài 2:

a: \(\left(x+y\right)^2+\left(x-y\right)^2\)

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

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

b: \(m^3+n^3+p^3-3mnp\)

\(=\left(m+n\right)^3-3mn\left(m+n\right)+p^3-3mnp\)

\(=\left(m+n\right)^3+p^3-3mn\left(m+n\right)-3mpn\)

\(=\left(m+n+p\right)\left[\left(m+n\right)^2-p\left(m+n\right)+p^2\right]-3mn\left(m+n+p\right)\)

\(=\left(m+n+p\right)\left(m^2+2mn+n^2-pm-pn+p^2-3mn\right)\)

\(=\left(m+n+p\right)\left(m^2+n^2+p^2-mn-mp-np\right)\)

Bài 2

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

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

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

\(=2x^2+2y^2\)

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

\(=VP\)

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

b) \(VP=\left(m+n+p\right)\left(m^2+n^2+p^2-mn-np-mp\right)\)

\(=m^3+mn^2+mp^2-m^2n-mnp-m^2p+m^2n+n^3+np^2-mn^2-n^2p-mnp+m^2p+n^2p+p^3-mnp-np^2-mp^2\)

\(=m^3+n^3+p^3+\left(mn^2-mn^2\right)+\left(mp^2-mp^2\right)+\left(-m^2n+m^2n\right)+\left(-mnp-mnp-mnp\right)+\left(n^2p-n^2p\right)+\left(np^2-np^2\right)+\left(m^2p-m^2p\right)\)

\(=m^3+n^3+p^3-3mnp\)

\(=VP\)

Vậy \(m^3+n^3+p^3-3mnp=\left(m+n+p\right)\left(m^2+n^2+p^2-mn-np-mp\right)\)

\(\left(a+b+c\right)^3=a^3+b^3+c^3\)

=>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=a^3+b^3+c^3\)

=>\(3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

=>\(\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

\(T=\left(a+b\right)\cdot\left(b+c\right)^2\cdot\left(c+a\right)^2\)

\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\cdot\left(b+c\right)\left(a+c\right)\)

\(=0\cdot\left(b+c\right)\left(a+c\right)\)

=0

a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt: \(x^2+x+1=y\), khi đó biểu thức trở thành:

\(y\left(y+1\right)-12\)

\(=y^2+y-12\)

\(=y^2-3y+4y-12\)

\(=y\left(y-3\right)+4\left(y-3\right)\)

\(=\left(y-3\right)\left(y+4\right)\)

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

\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(=\left(x^2-x+2x-2\right)\left(x^2+x+5\right)\)

\(=\left[x\left(x-1\right)+2\left(x-1\right)\right]\left(x^2+x+5\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)

b) \(\left(x^2+2x\right)^2+9x^2+18x+20\)

\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)

Đặt: \(x^2+2x=a\), khi đó biểu thức trở thành:

\(a^2+9a+20\)

\(=a^2+4a+5a+20\)

\(=a\left(a+4\right)+5\left(a+4\right)\)

\(=\left(a+4\right)\left(a+5\right)\)

\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)

c) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

Đặt: \(x^2+10x+20=y\), khi đó biểu thức trở thành:

\(\left(y-4\right)\left(y+4\right)+16\)

\(=y^2-16+16\)

\(=y^2\)

\(=\left(x^2+10x+20\right)^2\)

$\text{#}Toru$

$(x^2-2)^2+4(x-1)^2-4(x^2-2)(x-1)=0$

$\Leftrightarrow(x^2-2)^2-4(x^2-2)(x-1)+4(x-1)^2=0$

$\Leftrightarrow(x^2-2)^2-2\cdot(x^2-2)\cdot2(x-1)+[2(x-1)]^2=0$

$\Leftrightarrow[(x^2-2)-2(x-1)]^2=0$

$\Leftrightarrow(x^2-2-2x+2)^2=0$

$\Leftrightarrow(x^2-2x)^2=0$

$\Leftrightarrow x^2-2x=0$

$\Leftrightarrow x(x-2)=0$

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy: $x\in\{0;2\}$.

\(a,\left(x+1\right)^3-x\left(x-2\right)^2-1\\ =\left(x^3+3x^2+3x+1\right)-x\left(x^2-4x+4\right)-1\\ =x^3-x^3+3x^2+4x^2+3x-4x+1-1\\ =7x^2-x\\ ---\\ b,\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x^2-x+1\right)\\ =\left[\left(x+1\right)\left(x^2-x+1\right)\right].\left[\left(x-1\right)\left(x^2+x+1\right)\right]=\left(x^3+1\right)\left(x^3-1\right)=x^6-1\\ ---\\ c,\left(2x+1\right)^3=\left(2a\right)^3+3.\left(2a\right)^2.1+3.2a.1^2+1^3=8a^3+12a^2+6a+1\\ d,\left(3a-2b\right)^3=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\\ =27a^3-54a^2b+36ab^2-8b^3\)

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

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

\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)

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

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

\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)

`a, 8xy^2-2x^2y`

`= 2xy ( 4y - x)`

`b, x(x-y) -y(y-x)`

`= x(x-y) + y(x-y)`

`= (x-y)(x+y)`

`c, x(x-1) + (1-x)^2`

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

`= (x-1) (x+x-1)`

`=(x-1)(2x-1)`

a. $A=x(x-1)-y(x-1)$

$=(x-1)(x-y)$

Thay $x=2;y=1$ vào $A$, ta được:

$A=(2-1)(2-1)=1\cdot1=1$

Vậy $A=1$ tại $x=2;y=1$.

b. $B=x^5(x+2y)-x^3y(x+2y)+x^2y^2(x+2y)$

$=(x+2y)(x^5-x^3y+x^2y^2)$

Thay $x=10;y=-5$ vào $B$, ta được:

$B=[10+2\cdot(-5)][10^5-10^3\cdot(-5)+10^2\cdot(-5)^2]$

$=(10-10)(10^5+10^3\cdot5+10^2\cdot5^2)$

$=0$

Vậy $B=0$ tại $x=10;y=-5$.