Đáp án:
\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.\\
b)x = \dfrac{1}{2}\\
c)x = 1\\
d)x = 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0\\
\sqrt {x + 2\sqrt x + 1} = 3\\
\to \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} = 3\\
\to \left| {\sqrt x + 1} \right| = 3\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 3\\
\sqrt x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.\\
b)\sqrt {{{\left( {2x - 1} \right)}^2}} = 1 - 2x\\
\to \left| {2x - 1} \right| = 1 - 2x\\
\to \left[ \begin{array}{l}
2x - 1 = 1 - 2x\\
2x - 1 = - 1 + 2x
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = 2\\
- 1 = - 1\left( {ld} \right)
\end{array} \right.\\
\to x = \dfrac{1}{2}\\
c)DK:x \ge 1\\
\sqrt {{{\left( {\sqrt x - 1} \right)}^2}} = \sqrt x - 1\\
\to \left| {\sqrt x - 1} \right| = \sqrt x - 1\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = \sqrt x - 1\left( {ld} \right)\\
\sqrt x - 1 = - \sqrt x + 1
\end{array} \right.\\
\to 2\sqrt x = 2\\
\to x = 1\\
d)DK:x \ge 2\\
\sqrt {x - 2 - 2\sqrt {x - 2} .1 + 1} = \sqrt {x - 2} - 1\\
\to \sqrt {{{\left( {\sqrt {x - 2} - 1} \right)}^2}} = \sqrt {x - 2} - 1\\
\to \left| {\sqrt {x - 2} - 1} \right| = \sqrt {x - 2} - 1\\
\to \left[ \begin{array}{l}
\sqrt {x - 2} - 1 = \sqrt {x - 2} - 1\left( {ld} \right)\\
\sqrt {x - 2} - 1 = - \sqrt {x - 2} + 1
\end{array} \right.\\
\to 2\sqrt {x - 2} = 2\\
\to \sqrt {x - 2} = 1\\
\to x - 2 = 1\\
\to x = 3
\end{array}\)
