Question

\(\sqrt{x^2+4}=\sqrt{2x+3}\) . Tìm x

Answer 1

Lời giải :

ĐK: \(x\ge\frac{-3}{2}\)

\(pt\Leftrightarrow x^2+4=2x+3\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\) ( thỏa )

Vậy....

Answer 2

$\sqrt{x^2+4}=\sqrt{2x+3}$ ( 1)

Đk: \(x\ge-\frac{3}{2}\)

(1) <=> \(x^2+4=2x+3\)

<=> \(x^2-2x+4-3=0\)

<=> \(\left(x-1\right)^2=0\)

<=> x =1 ( tmđkxđ )

vậy S = \(\left\{1\right\}\)