a. Ta có : \(4x^2-6x+9=4x^2-6x+\dfrac{9}{4}+\dfrac{27}{4}\)

\(=\left[\left(2x\right)^2-6x+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{27}{4}\)

\(=\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)

Vì \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\)

nên \(\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\forall x\)

b.Ta có : \(x^2+2y^2-2xy+y+1=\left(x^2+y^2-2xy\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

\(=\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

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

\(\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\)

nên \(\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\forall x;y\)

\(x^2+2xy+2y^2+y+\frac{1}{2}\)

\(=x^2+2xy+y^2+y^2+y+\frac{1}{2}\)

\(=\left(x+y\right)^2+y^2+y+\frac{1}{2}\)

Có : \(\left(x+y\right)^2\ge0\)

\(y^2\ge y\ge0\Rightarrow y^2+y\ge0\)

\(\frac{1}{2}>0\)

\(\Rightarrow x^2+2xy+2y^2+y+\frac{1}{2}>0\) với mọi x y 

Ta có

\(x^2+2xy+2y^2+y+\frac{1}{2}\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)

\(=\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{2}\)

Mà \(\begin{cases}\left(x^2+2xy+y^2\right)\ge0\\\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)\ge0\\\frac{1}{4}>0\end{cases}\)

\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}>0\)

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

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

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

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

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

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

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

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

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

\(=\left(x-y+1\right)^2+\left(y-1\right)^2+3>0\forall x;y\)

\(A=2xy^2\cdot\dfrac{1}{2}x^2y^2x=x^4y^4>0\)(Vì x,y khác 0)

\(x^2-2xy-x+1+2y^2=x^2-x\left(2y+1\right)+\frac{\left(2y+1\right)^2}{4}-\frac{\left(2y+1\right)^2}{4}+2y^2+1\)

\(=\left(x-\frac{2y+1}{2}\right)^2+\frac{1}{4}\left(2y-1\right)^2+\frac{1}{2}>0\)

bn có thể lm rõ hơn dc chứ

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

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

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

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

<=> ( x²+2xy+y²+2x+2y+1)+(x²-4x+4)-3

=( x+y+1)²+(x-2)² -3 >= -3

dấu = xảy ra <=> x=2 và y=-3