Question

a)\(\sqrt{9+x}=5-\sqrt{4+2x}\)

b)\(\sqrt{8+3x}-\sqrt{5+3x}=\sqrt{5x-4}-\sqrt{5x-7}\)

Answer 1

a/ ĐKXĐ: \(x\ge-2\)

\(\Leftrightarrow\sqrt{x+9}-3+\sqrt{2x+4}-2=0\)

\(\Leftrightarrow\frac{x}{\sqrt{x+9}+3}+\frac{2x}{\sqrt{2x+4}+2}=0\)

\(\Leftrightarrow x\left(\frac{1}{\sqrt{x+9}+3}+\frac{2}{\sqrt{2x+4}+2}\right)=0\)

\(\Leftrightarrow x=0\)

b/ ĐKXĐ: \(x\ge\frac{7}{5}\)

\(\Leftrightarrow\sqrt{5x-7}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{3x+8}\)

\(\Leftrightarrow\frac{2x-12}{\sqrt{5x-7}+\sqrt{3x+5}}=\frac{2x-12}{\sqrt{5x-4}+\sqrt{3x+8}}\)

\(\Leftrightarrow2x-12=0\Rightarrow x=6\)