\(\sqrt{x+1}+\sqrt{4-x}-\sqrt{\left(x+1\right)\left(4-x\right)}=1\)
⇔ \(\sqrt{x+1}+\sqrt{4-x}-\sqrt[]{4x-x}^2+4-x=1\)
⇔ \(\sqrt{x+1}+\sqrt{4-x}-\sqrt{3x-x^2+4}=1\)
⇔ \(\sqrt{x+1}+\sqrt{4-x}=1+\sqrt{3x-x^2+4}\)
⇔ \(x+1+2\sqrt{\left(x+1\right)\left(4-x\right)}+4-x=1+2\sqrt{3x-x^2+4}+3x-x^2+4\)
⇔ \(1+2\sqrt{\left(x+1\right)\left(4-x\right)}+4=1+2\sqrt{3x-x^2+4}+3x-x^2+4\)
⇔ \(2\sqrt{4x-x^2+4-x}=2\sqrt{3x-x^2+4}+3x+-x^2\)
⇔ \(2\sqrt{3x-x^2+4}=2\sqrt{3x-x^2+4}+3x-x^2\)
⇔\(-3x+x^2=0\)
⇔ \(-x\left(3-x\right)=0\)
⇔ -x = 0 hoặc 3 - x = 0
⇔ x = 0 hoặc x = 3
Vậy x = 0 hoặc x = 3
ĐKXĐ: \(-1\le x\le4\)
\(\sqrt{x+1}-1+\sqrt{4-x}-\sqrt{\left(x+1\right)\left(4-x\right)}=0\)
\(\Leftrightarrow\sqrt{x+1}-1-\sqrt{4-x}\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-1\right)\left(1-\sqrt{4-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{4-x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\4-x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
