Question

[Thanh Trần Nhật]

Giải hệ phương trình:

\(\left\{{}\begin{matrix}2x^2+y^2-3xy-4x+3y+2=0\\\sqrt{x^2-y+3}+\sqrt{y-x+1}=2\end{matrix}\right.\)

Answer 1

[Lightning Farron]

\(\sqrt{x^2-y+3}+\sqrt{y-x+1}=2\)

Xét \(pt\left(1\right)\Leftrightarrow2x^2+y^2-3xy-4x+3y+2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=2x-2\end{matrix}\right.\)

*)\(y=x-1\) thay vao \(pt(2)\) :

\(pt\Leftrightarrow\sqrt{x^2-x+4}=2\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=0\end{matrix}\right.\)

*)\(y=2x-2\) thay vao \(pt(2)\):

\(pt\Leftrightarrow\sqrt{x^2-2x+5}+\sqrt{x-1}=2\)

\(\Leftrightarrow\dfrac{x^2-2x+1}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x-1}{\sqrt{x^2-2x+5}+2}+\dfrac{1}{\sqrt{x-1}}\right)=0\)

\(\Leftrightarrow x=1\)\(\Leftrightarrow y=0\)

Answer 2

[Phan Thế Nghĩa]

sai r bạn ơi!