Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\sqrt {{x^2} + 2x + 1}  = 0\\
 \Leftrightarrow \sqrt {{{\left( {x + 1} \right)}^2}}  = 0\\
 \Leftrightarrow \left| {x + 1} \right| = 0\\
 \Leftrightarrow x =  - 1\\
2,\\
x - 4\sqrt x  + 3 = 0\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
 \Leftrightarrow \left( {x - \sqrt x } \right) - \left( {3\sqrt x  - 3} \right) = 0\\
 \Leftrightarrow \sqrt x \left( {\sqrt x  - 1} \right) - 3.\left( {\sqrt x  - 1} \right) = 0\\
 \Leftrightarrow \left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  - 1 = 0\\
\sqrt x  - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 1\\
\sqrt x  = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 9
\end{array} \right.\\
4,\\
\sqrt {\dfrac{{9\left( {x - 3} \right)}}{{16}}}  - 5.\sqrt {\dfrac{{x - 3}}{{25}}}  = \sqrt {9\left( {x - 3} \right)}  + 1\,\,\,\,\,\,\,\,\,\,\left( {x \ge 3} \right)\\
 \Leftrightarrow \sqrt {\dfrac{9}{{16}}.\left( {x - 3} \right)}  - 5.\sqrt {\dfrac{1}{{25}}\left( {x - 3} \right)}  = \sqrt {9.\left( {x - 3} \right)}  + 1\\
 \Leftrightarrow \dfrac{3}{4}.\sqrt {x - 3}  - 5.\dfrac{1}{5}\sqrt {x - 3}  = 3.\sqrt {x - 3}  + 1\\
 \Leftrightarrow \dfrac{3}{4}\sqrt {x - 3}  - \sqrt {x - 3}  = 3\sqrt {x - 3}  + 1\\
 \Leftrightarrow  - \dfrac{{13}}{4}\sqrt {x - 3}  = 1\\
 \Leftrightarrow \sqrt {x - 3}  =  - \dfrac{4}{{13}}\,\,\,\,\,\left( {vn} \right)\\
5,\\
3\sqrt x  - 2 < 0\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
 \Leftrightarrow \sqrt x  < \dfrac{2}{3}\\
 \Leftrightarrow 0 \le x < \dfrac{4}{9}\\
6,\\
\dfrac{{\sqrt {x + 1} }}{{\sqrt x }} > 2\,\,\,\,\,\,\,\,\,\,\,\left( {x > 0} \right)\\
 \Leftrightarrow \sqrt {x + 1}  > 2\sqrt x \\
 \Leftrightarrow x + 1 > 4x\\
 \Leftrightarrow 1 > 3x\\
 \Leftrightarrow 0 < x < \dfrac{1}{3}
\end{array}\)

Em xem lại đề câu 3 nhé!