1/ Đặt \(\sqrt{5x-x^2}=a\ge0\)
Thì ta có:
\(a-2a^2+6=0\)
\(\Leftrightarrow\left(2-a\right)\left(2a+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-\dfrac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{5x-x^2}=2\)
\(\Leftrightarrow x^2-5x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y+xy=3\\\sqrt{x}+\sqrt{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+\sqrt{y}\right)^2+xy-2\sqrt{xy}=3\left(1\right)\\\sqrt{x}+\sqrt{y}=2\left(2\right)\end{matrix}\right.\)
\(\Rightarrow\left(1\right)\Leftrightarrow xy-2\sqrt{xy}+1=0\)
\(\Leftrightarrow\sqrt{xy}=1\)
\(\Leftrightarrow\sqrt{y}=\dfrac{1}{\sqrt{x}}\) thế vô (2) ta được
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}=2\)
\(\Leftrightarrow x-2\sqrt{x}+1=0\)
\(\Rightarrow x=1\)
\(\Rightarrow y=1\)
Cách khác cách của anh Hung Nguyen
\(\sqrt{5x-x^2}+2x^2-10x+6=0\)
\(\Rightarrow\sqrt{5x-x^2-\dfrac{25}{4}+\dfrac{25}{4}}+2x^2-10x+6=0\)
\(=\sqrt{5x+\left(-x^2\right)+\dfrac{25}{4}-\dfrac{25}{4}}+2x^2-10x+6=0\)
\(\Rightarrow\sqrt{\left(-x+-\dfrac{5}{2}\right)^2-\dfrac{25}{4}}+2x^2-10x+6=0\)
\(\Rightarrow\left|-x+-\dfrac{5}{2}\right|-\dfrac{5}{2}+2x^2-10x+6=0\)
\(\Rightarrow x-\dfrac{5}{2}-\dfrac{5}{2}+2x^2-10x+6=0\)
\(\Rightarrow x-5+2x^2-10x+6=0\)
\(\Rightarrow-9x+1+2x^2=0\)
Đến đây em bí r ạ:3
