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Question

giải phương trình $\left\{{}\begin{matrix}x^2-xy-x+3y-6=0\\sqrt{5x-6}+\sqrt{16-3y}=2x^2-2x+y-4\end{matrix}\right.$

Nguyễn Huy Thắng · Accepted Answer

$\left(1\right)\Leftrightarrow\left(x-3\right)\left(x-y+2\right)=0$

+)$x=3$ thi $\left(2\right)\Leftrightarrow\sqrt{5\cdot3-6}+\sqrt{16-3y}=2\cdot3^2-2\cdot3+y-4$

$\Leftrightarrow\sqrt{16-3y}=y+5$$\Rightarrow y^2+13y+9=0$$\Rightarrow y=\frac{\sqrt{133}-13}{2}$

+)$y=x+2$ thi $\sqrt{5x-6}+\sqrt{16-3\cdot\left(x+2\right)}=2x^2-2x+x+2-4$

$\Leftrightarrow\sqrt{10-3x}+\sqrt{5x-6}=2x^2-x-2$

$\Rightarrow x=2;y=4$Câu trả lời: $\left(1\right)\Leftrightarrow\left(x-3\right)\left(x-y+2\right)=0$

+)$x=3$ thi $\left(2\right)\Leftrightarrow\sqrt{5\cdot3-6}+\sqrt{16-3y}=2\cdot3^2-2\cdot3+y-4$

$\Leftrightarrow\sqrt{16-3y}=y+5$$\Rightarrow y^2+13y+9=0$$\Rightarrow y=\frac{\sqrt{133}-13}{2}$

+)$y=x+2$ thi $\sqrt{5x-6}+\sqrt{16-3\cdot\left(x+2\right)}=2x^2-2x+x+2-4$

$\Leftrightarrow\sqrt{10-3x}+\sqrt{5x-6}=2x^2-x-2$

$\Rightarrow x=2;y=4$

Violympic toán 9