Question

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

Answer 1

PT <=> \(\left(\sqrt{2x+3}\right)^2=\left(1+\sqrt{2}\right)^2\)

<=> \(2x+3=3+2\sqrt{2}\)

<=> \(2x-2\sqrt{2}=0\)

<=> \(2\left(x-\sqrt{2}\right)=0\)

<=> \(x=\sqrt{2}\)

KL: \(x\in\left\{\sqrt{2}\right\}\)

Answer 2

Điều kiện: x\(\ge-\frac{3}{2}\)

Có: \(\sqrt{2x+3}=1+\sqrt{2}\)

\(\Leftrightarrow\:2x+3=1+2\sqrt{2}+4\)

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

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