Question

\(\sqrt{2x+2\sqrt{2x-1}}\)+\(\sqrt{2x+3+4\sqrt{2x-1}}\)= 6

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Answer 1

Lời giải:

ĐK: $x\geq \frac{1}{2}$

\(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x+3+4\sqrt{2x-1}}=6\)

\(\Leftrightarrow \sqrt{(2x-1)+2\sqrt{2x-1}+1}+\sqrt{(2x-1)+4\sqrt{2x-1}+4}=6\)

\(\Leftrightarrow \sqrt{(\sqrt{2x-1}+1)^2}+\sqrt{(\sqrt{2x-1}+2)^2}=6\)

\(\Leftrightarrow \sqrt{2x-1}+1+2\sqrt{2x-1}+2=6\)

\(\Leftrightarrow 2\sqrt{2x-1}=3\Leftrightarrow \sqrt{2x-1}=\frac{3}{2}\)

\(\Rightarrow 2x-1=\frac{9}{4}\Rightarrow x=\frac{13}{8}\) (thỏa mãn)