\(ĐK:0\le x\le5\)
\(\Leftrightarrow\left(\sqrt{x}+\sqrt{5-x}\right)^2=10x-2x^2+1\)
\(\Leftrightarrow x+2\sqrt{x\cdot\left(5-x\right)}+5-x=10x-2x^2+1\)
\(\Leftrightarrow2\sqrt{5x-x^2}+5-10x+2x^2-1=0\)
\(\Leftrightarrow2\sqrt{5x-x^2}+4+2x^2-10x=0\)
\(\Leftrightarrow\sqrt{5x-x^2}+2+x^2-5x=0\)
\(\Leftrightarrow5x-x^2-\sqrt{5x-x^2}-2=0\)
Đặt \(\sqrt{5x-x^2}=a\left(a\ge0\right)\) ta có:
\(a^2-a-2=0\)
\(\Leftrightarrow\left(a+1\right)\left(a-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-1\left(L\right)\\a=2\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{5x-x^2}=2\)
\(\Leftrightarrow5x-x^2=4\)
\(\Leftrightarrow x^2-5x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)(T/m)
Vậy \(S=\left\{1;4\right\}\)
đk \(\left\{{}\begin{matrix}0\le x\le5\\-2x^2+10x+1\ge0\end{matrix}\right.\)
Ta có \(\sqrt{x}+\sqrt{5-x}=\sqrt{2x\left(5-x\right)+1}\)
\(\Leftrightarrow5+2\sqrt{x\left(5-x\right)}=2x\left(5-x\right)+1\)
Đặt \(\sqrt{x\left(5-x\right)}=a\left(a\ge0\right)\)
\(\Leftrightarrow2a^2-2a-4=0\Leftrightarrow a^2-a-2=0\Leftrightarrow a=2;a=-1\left(l\right)\)
Với a = 2
\(x\left(5-x\right)=4\Leftrightarrow x^2-5x+4=0\Leftrightarrow x=4;x=1\)(tm)
