Question

Giải phương trình:

\(\sqrt{1-x}+\sqrt{4+x}=3\)

Answer 1

\(ĐK:-4\le x\le1\\ PT\Leftrightarrow\sqrt{1-x}-1+\sqrt{x+4}-2=0\\ \Leftrightarrow\dfrac{1-x-1}{\sqrt{1-x}+1}+\dfrac{x+4-4}{\sqrt{x+4}+2}=0\\ \Leftrightarrow\dfrac{x}{\sqrt{x+4}+2}-\dfrac{x}{\sqrt{1-x}+1}=0\\ \Leftrightarrow x\left(\dfrac{1}{\sqrt{x+4}+2}-\dfrac{1}{\sqrt{1-x}+1}=0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\\dfrac{1}{\sqrt{x+4}+2}=\dfrac{1}{\sqrt{1-x}+1}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{x+4}+2=\sqrt{1-x}+1\\ \Leftrightarrow\sqrt{x+4}+\sqrt{1-x}=-1\)

Vì \(\sqrt{x+4}+\sqrt{1-x}\ge0>-1\)

Do đó (1) vô nghiệm

Vậy PT có nghiệm x=0