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

\(\Rightarrow x^2-2xy+y^2+2\left(x-y\right)+1+y^2-2y+1=0\)

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

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

=>................

(1)

(x+1)(x-7)+17>0

<=>x^2-6x+9+1>0

<=>(x-3)^2+1>0(dpcm)

..

(7)

-y^2+4y-4-|x+1|≤0

<=>-(y-2)^2-|x+1|≤0

sum 2 so khong duong ko the la so (+)=>dpcm

1.(x+1)(x-7)+17=(x-3)²+1>0

2.-20-(x-5)(x+3)=-34-(x-1)²<0

3.-2(x+3)-(x-2)(x+2)=-(x+1)²-1<0

4.x²+y²+2x+2y+3=(x+1)²+(y+1)²+1>0

5.2x²+2x+y²+2y+5=2(x+1/2)²+(y+1)²+2>0

6.2x²+2y²+2xy+2x+4y+6=(x+y)²+(x+1)²+(y+2)²+1>0

7.-y²+4y-4-/x+1/=-(y-2)²-/x+1/≤0

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

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

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

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

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

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

Vì \(\left(x-y+1\right)^2+\left(y-1\right)^2\ge0\forall x;y\)

Nên \(\left(x-y+1\right)^2+\left(y-1\right)^2+1>0\forall x;y\)

Vậy \(x^2+2y^2-2xy+2x-4y+3>0\forall x;y\)

Bài 1:

A=x² +y² -2x-2y+2xy+5

=x² +y² -2x-2y+2xy+1+4

=xy+x²-x+xy+y²-y-y-x+1+4

=x(x+y-1)+y(x+y-1)-1(x+y-1)

=(x+y-1)(x+y-1)

=(x+y-1)²+4.Với x+y=3

=>A=(3-1)²+4=2²+4=8

Bài 2:

B=x^2 +4y^2-2x-4y-4xy+10

=-2xy+x²-x-2xy+4y²+2y-x+2y+1-8y+9

=x(x-2y-1)-2y(x-2y-1)-1(x-2y-1)-8y+9

=(x-2y-1)(x-2y-1)-8y+9

=(x-2y-1)²-8y+9

Với x-2y=5.Ta có:... tự thay

Bài 3: chịu

\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)

\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)

ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)

Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)

T i c k cho mình 1 cái nha mới bị trừ 50 đ