\(đk:-5\le x\le3\)
\(\sqrt{x+5}+\sqrt{3-x}=2\left(\sqrt{15-2x-x^2}+1\right)\) (1)
\(\Leftrightarrow\sqrt{x+5}+\sqrt{3-x}=2\left(\sqrt{\left(x+5\right)\left(3-x\right)}+1\right)\)
\(\Leftrightarrow\sqrt{x+5}=\sqrt{3-x}=2\sqrt{\left(x+5\right)\left(3-x\right)}+2\)
đặt \(\sqrt{x+5}+\sqrt{3-x}=t\) (đk t > 0)
\(\Leftrightarrow t^2=x+5+2\sqrt{\left(x+5\right)\left(3-x\right)}+3-x\)
\(\Leftrightarrow t^2=8+2\sqrt{\left(x+5\right)\left(3-x\right)}\)
\(\Leftrightarrow t^2=6+\left(2+2\sqrt{\left(x+5\right)\left(3-x\right)}\right)\) và (1)
\(\Rightarrow t=t^2-6\)
\(\Leftrightarrow t^2-t-6=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=-2\left(loai\right)\\t=3\end{cases}}\)
ta có : \(8+2\sqrt{\left(x+5\right)\left(3-x\right)}=3^2=9\)
\(\Leftrightarrow2\sqrt{\left(x+5\right)\left(3-x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(x+5\right)\left(3-x\right)}=\frac{1}{2}\)
\(\Leftrightarrow\left(x+5\right)\left(3-x\right)=\frac{1}{4}\)
\(\Leftrightarrow59-8x-4x^2=0\)
\(\Leftrightarrow4x^2+8x+4-63=0\)
\(\Leftrightarrow4\left(x+1\right)^2=63\) \(\Leftrightarrow\left(x+1\right)^2=\frac{63}{4}\Leftrightarrow x+1=\pm\sqrt{\frac{63}{4}}\)
\(\Leftrightarrow x=\pm\sqrt{\frac{63}{4}}-1\left(tm\right)\)
