Đáp án:
\(\left[ \begin{array}{l}x=0 \\x=4 \end{array} \right.\)
Giải thích các bước giải:
$\sqrt[]{2x+1}$-2$\sqrt[]{x}$+1=0(1)
ĐKXĐ: `2x+1>=0`; `x>=0<=>x>=0 `
`(1)`<=> $\sqrt[]{2x+1}$=2$\sqrt[]{x}$-1
`<=>2x+1=4x-4`$\sqrt[]{x}$`+1`
`<=> 4`$\sqrt[]{x}$`=2x`
`<=>2`$\sqrt[]{x}$`=x`
`<=> 4x=x^2`
`<=> x^2-4x=0`
`<=> x(x-4)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x-4=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0 (TM)\\x=4 (TM)\end{array} \right.\)
