6x² - 26x - 6y² + 39y - 5xy - 5 = 0

<=> (6x² - 9xy) + (4xy - 6y²) + ( - 26x + 39y) = 5

<=> (2x - 3y)(3x + 2y - 13) = 5

Tới đây tự làm nốt nhé

\(2x^2+7y^2+3x-6y=5xy-7\)

\(\Leftrightarrow x^2-5xy+\frac{25}{4}y^2+3x-\frac{15}{2}y+\frac{9}{4}+\frac{3}{4}y^2+\frac{3}{2}y+\frac{3}{4}+x^2+4=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}y\right)^2+2.\left(x-\frac{5}{2}y\right).\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{3}{4}\left(y^2+2y+1\right)+x^2+4=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}y+\frac{3}{2}\right)^2+\frac{3}{4}\left(y+1\right)^2+x^2+4=0\)

Thấy ngay \(VT>0\)

=> Pt vô nghiệm 

Sure ?

\(2x^2+7y^2+3x-6y=5xy-7\)

<=> \(16x^2+56y^2+24x-48y=40xy-56\)

<=> \(\left(16x^2-40xy+25y^2\right)+6\left(4x-5y\right)+9+\left(31y^2-18y+47\right)=0\)

<=> \(\left(16x^2-40xy+25y^2\right)+6\left(4x-5y\right)+9+\left(31y^2-18y+47\right)=0\)

<=> \(\left(4x-5y\right)^2+6\left(4x-5y\right)+9+\left(31y^2-18y+47\right)=0\)

<=> \(\left(4x-5y+3\right)^2+\left(31y^2-18y+47\right)=0\)(1)

Mà \(31y^2-18y+47>0\)với mọi y 

=> (1) vô nghiệm

\(x^2-4x+y^2-6y+15=0\)

\(\Rightarrow\left(x^2-4x+4\right)+\left(y^2-6y+9\right)+2=0\)

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

Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)

              \(\left(y-3\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-2\right)^2+\left(y-3\right)^2\ge0\forall x;y\)

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

\(\Rightarrow\)Phương trình vô nghiệm.

\(x^2-4x+y^2-6y+15=0\)

\(\Leftrightarrow x^2-4x+4+y^2-6y+9+2=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2-6y+9\right)+2=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(y-3\right)^2+2=0\)

Mà:  

\(\left(x-2\right)^2\ge0\forall x\)

\(\left(y-3\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-2\right)^2+\left(y-3\right)^2+2\ge2\forall x,y\)

\(\Rightarrow\left(x-2\right)^2+\left(y-3\right)^2+2=0\) (vô lý)

⇒ Phương trình vô nghiệm:

\(x\in\varnothing\)

Lời giải:

$x^2-5xy+6y^2=0$

$\Leftrightarrow x^2-2xy-3xy+6y^2=0$

$\Leftrightarrow x(x-2y)-3y(x-2y)=0$

$\Leftrightarrow (x-2y)(x-3y)=0$

$\Rightarrow x=2y$ hoặc $x=3y$

Nếu $x=2y$. Thay vào PT $(2)$ ta được:

$4(2y)^2+2.2y.y+6y-27=0$

$\Leftrightarrow 20y^2+6y-27=0$

$\Leftrightarrow y=\frac{-3+3\sqrt{61}}{20}$

$\Rightarrow x=\frac{-3+3\sqrt{61}}{10}$

Nếu $x=3y$. Thay vào PT $(2)$ ta được:

$4(3y)^2+2.3y.y+6y-27=0$

$\Leftrightarrow 42y^2+6y-27=0$

$\Leftrightarrow y=\frac{-1\pm \sqrt{127}}{14}$

$\Rightarrow x=\frac{-3\pm 3\sqrt{127}}{14}$