\(Q=x^2+2y^2+2z^2+2xy-2yz-2xz-2y+4z+5=\left[\left(x^2+2xy+y^2\right)-2z\left(x+y\right)+z^2\right]+\left(y^2-2y+1\right)+\left(z^2+4z+4\right)=\left(x+y-z\right)^2+\left(y-1\right)^2+\left(z+2\right)^2\ge0\)

\(minQ=0\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-3\\y=1\\z=-2\end{matrix}\right.\)

`Q=x^2+2y^2+2z^2+2xy-2yz-2xz-2y+4z+5`

`Q=(x^2+y^2-z^2+2xy-2yz-2xz)+(y^2-2y+1)+(z^2+4z+4)`

`Q=(x+y-z)^2+(y-1)^2+(z+2)^2`

Ta thấy :

`(x+y-z)^2>=0`

`(y-1)^2>=0`

`(z+2)^2>=0`

`=>(x+y-z)^2+(y-1)^2+(z+2)^2>=0`

Dấu = xảy ra 

`<=>` $\begin{cases}x+y-z=0\\y-1=0\\z+2=0\end{cases}$

`<=>` $\begin{cases}x=-3\\y=1\\z=-2\end{cases}$

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2z+1\right)< 1\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-1\right)^2< 1\)

Nếu tồn tại 1 trong 3 số \(x-y;y-z;z-1\) khác 0

Do x; y; z nguyên

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge1\) (vô lý)

\(\Rightarrow x-y=y-z=z-1=0\)

\(\Leftrightarrow x=y=z=1\)

\(P=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+\left(y^2-8y+16\right)-16\\ P=\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-4\right)^2-16\\ P=\left(x-y+1\right)^2+\left(y-4\right)^2-16\ge-16\)

\(P_{min}=-16\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

\(P=\left(x^2+y^2+1-2xy+2x-2y\right)+\left(y^2-8y+16\right)-16\\ =\left(x-y+1\right)^2+\left(y-4\right)^2-16\\ \ge-16\)

dấu = xảy ra khi và chỉ khi y=4,x=3

a, b, nhân vào là ra à

c, nghe cứ là lạ

d, cũng nhân là ra hà

\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5\)

a) Ta có: \(VT=\left(x-y-z\right)^2\)

\(=\left(x-y-z\right)\left(x-y-z\right)\)

\(=x^2-xy-xz-yx+y^2+yz-zx+zy+z^2\)

\(=x^2+y^2+z^2-2xy+2yz-2xz\)

=VP(đpcm)

b) Ta có: \(VT=\left(x+y-z\right)^2\)

\(=\left(x+y-z\right)\left(x+y-z\right)\)

\(=x^2+xy-xz+yx+y^2-yz-zx-zy+z^2\)

\(=x^2+y^2+z^2+2xy-2yz-2zx\)

=VP(đpcm)

c) Sửa đề: Chứng minh \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4-y^4\)

Ta có: \(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)

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

\(=x^4-y^4\)

=VP(đpcm)

d) Ta có: \(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)

\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)

\(=x^5+y^5\)

=VP(đpcm)

\(P=\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\)
\(P=\frac{\left[\left(\frac{x}{\sqrt{x^2+2yz}}\right)^2+\left(\frac{y}{\sqrt{y^2+2xz}}\right)^2+\left(\frac{z}{\sqrt{z^2+2xy}}\right)^2\right]\left[\sqrt{x^2+2yz}^2+\sqrt{y^2+2xz}^2+\sqrt{z^2+2xy}^2\right]}{x^2+2yz+y^2+2xz+z^2+2xy}\)

\(P\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)(Bunyakovski)

Dấu "=" xảy ra <=> \(\frac{x}{x^2+2yz}=\frac{y}{y^2+2xz}=\frac{z}{z^2+2xy}\Leftrightarrow x=y=z\)

Vậy GTNN P=1 <=> x=y=z

Ngay ở trên hai cái [...] [...] nhân với nhau ấy, tại nó dài quá 

toàn lớp 8e trường trung học cơ sở đan phượng đẹp trai nhất hanhdf tinh đêyy