Question

Tìm X

\(\sqrt{x+5}=1+\sqrt{x}\)             

 

Answer 1

ĐK : $x ≥ 0$

$\sqrt{x + 5} = 1 + \sqrt{x}$
$⇔ x + 5 = 1 + 2\sqrt{x} + x$
$⇔ 2 \sqrt{x} = 4$
$⇔ \sqrt{x} = 2$
$⇔ x = 4$ (thoả mãn)

Answer 2

`\sqrt{x+5}``=` `1 +\sqrt{x}`

\(\text{ĐK}:x\ge0\)

\(PT\left(1\right)\Leftrightarrow\sqrt{x+5}^2=\left(1+\sqrt{x}\right)^2\)

\(\Leftrightarrow x+5=1^2+\sqrt{x}^2+2\sqrt{x}\)

\(\Leftrightarrow x+5=1+x+2\sqrt{x}\)

\(\Leftrightarrow x+5-1-x=2\sqrt{x}\)

\(\Leftrightarrow4=2\sqrt{x}\)

\(\Leftrightarrow2=\sqrt{x}\)

\(\Leftrightarrow x=4\left(TM\right)\)