rgthaegƯ mk chỉ giải được phần a thui 

  x^2 + 2y^2 - 2xy + 2x + 2 - 4y =0 
<=>x^2 + y^2 - 2xy+2x-2y+y^2-2y+1+1=0 
<=>(x-y)^2+2(x-y)+1+(y-1)^2=0 
<=>(x-y+1)^2+(y-1)^2=0 
<=>y=1;x=0

a)\(2x^2+3x+5=0\)

\(\Leftrightarrow4x^2+6x+10=0\)

\(\Leftrightarrow\left(2x\right)^2+2.2x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{31}{4}=0\)

\(\Leftrightarrow\left(2x+\dfrac{3}{2}\right)^2=-\dfrac{31}{4}\left(vn\right)\)

b) PT \(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=-1\left(vn\right)\) ( do \(VT\ge0\forall x,y\) )

c) PT \(\Leftrightarrow\left(x^2-2xy+y^2\right)+y^2+2x-6y+10=0\)

\(\Leftrightarrow\left(x-y\right)^2+2\left(x-y\right)+1+y^2-4y+4+5=0\)

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

Vậy PT vô nghiệm

a: 2x^2+3x+5=0

=>x^2+3/2x+5/2=0

=>x^2+2*x*3/4+9/16+31/16=0

=>(x+3/4)^2+31/16=0(vô lý)

b: x^2-2x+y^2-4y+6=0

=>x^2-2x+1+y^2-4y+4+1=0

=>(x-1)^2+(y-2)^2+1=0(vô lý)

x²+4y²-2x+4y+2=0

<=>x²-2x+1+4y²+4y+1=0

<=>(x-1)²+(2y+1)²=0

<=>x-1=0 và 2y+1=0

<=>x=1 và y=-1/2

Ta có:

\(x^2-2xy+2y^2-2x+6y+5=\left(x^2-xy+y^2\right)+y^2-2\left(x-y\right)+4y+5\)

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

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

\(\Rightarrow\hept{\begin{cases}x-y=1\\y=-2\end{cases}\Rightarrow\hept{\begin{cases}x=y+1=-1\\y=-2\end{cases}}}\)

           \(x^2-2xy+2y^2-2x+6y+5=0\)

\(\Leftrightarrow\)\(x^2-2x\left(y+1\right)+\left(y^2+2y+1\right)+\left(y^2+4y+4\right)=0\)

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

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

\(\Leftrightarrow\)\(\hept{\begin{cases}x-y-1=0\\y+2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-1\\y=-2\end{cases}}\)

\(\frac{ }{ }\)

\(x^2-2xy+2y^2-2x+6y+5=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)-2\left(x-y\right)+1+y^2+4y+4=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\y+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=-2\end{cases}}\)

\(x^2+2y^2-3xy+2x-4y+3=0\)

\(\Leftrightarrow\left(x^2-3xy+\frac{9}{4}y^2\right)+2\left(x-\frac{3}{2}y\right)+1-\left(\frac{1}{4}y^2+y+1\right)+3=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}y\right)^2+2\left(x-\frac{3}{2}y\right)+1-\left(\frac{1}{2}y+1\right)^2+3=0\)

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

\(\Leftrightarrow\left(x-\frac{3}{2}y+1-\frac{1}{2}y-1\right)\left(x-\frac{3}{2}y+1+\frac{1}{2}y+1\right)=-3\)

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

Đến đây tự làm ( Dễ )