Question

[Chan Chan]

1) Giải phương trình:\(\sqrt{5x-x^2}+2x^2-10x+6=0\)

2) Giải hệ phương trình:\(\left\{{}\begin{matrix}x+y+xy=3\\\sqrt{x}+\sqrt{y}=2\end{matrix}\right.\)

Answer 1

[Hung nguyen]

1/ Đặt \(\sqrt{5x-x^2}=a\ge0\)

Thì ta có:

\(a-2a^2+6=0\)

\(\Leftrightarrow\left(2-a\right)\left(2a+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-\dfrac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{5x-x^2}=2\)

\(\Leftrightarrow x^2-5x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Answer 2

[Hung nguyen]

\(\left\{{}\begin{matrix}x+y+xy=3\\\sqrt{x}+\sqrt{y}=2\end{matrix}\right.\)

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

\(\Rightarrow\left(1\right)\Leftrightarrow xy-2\sqrt{xy}+1=0\)

\(\Leftrightarrow\sqrt{xy}=1\)

\(\Leftrightarrow\sqrt{y}=\dfrac{1}{\sqrt{x}}\) thế vô (2) ta được

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}=2\)

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

\(\Rightarrow x=1\)

\(\Rightarrow y=1\)