\(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\) \(\left(4\ge x\ge-1\right)\)
Đặt: \(a=\sqrt{x+1}+\sqrt{4-x}\) \(\left(a\ge0\right)\)
\(\Rightarrow a^2=x+1+4-x+2\sqrt{\left(x+1\right)\left(4-x\right)}\)
\(a^2=5+2\sqrt{\left(x+1\right)\left(4-x\right)}\)
\(\Rightarrow\dfrac{a^2-5}{2}=\sqrt{\left(x+1\right)\left(4-x\right)}\)
\(\Leftrightarrow a+\dfrac{a^2-5}{2}=5\)
\(\Leftrightarrow\left[{}\begin{matrix}a=3\left(n\right)\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x+1}+\sqrt{4-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x+1\right)\left(4-x\right)}=9\)
\(\Leftrightarrow2\sqrt{\left(x+1\right)\left(4-x\right)}=4\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=2\)
\(\Leftrightarrow4x-x^2+4-x=4\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=3\left(n\right)\end{matrix}\right.\)
