Đáp án:
\(\begin{array}{l}
c,\\
x = 7\\
d,\\
\left[ \begin{array}{l}
x = \dfrac{{11}}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
e,\\
\left[ \begin{array}{l}
x = 4\\
x = - \dfrac{8}{3}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
\sqrt 2 x - 3\sqrt {18} + 2\sqrt {72} = \sqrt {200} \\
\Leftrightarrow \sqrt 2 x - 3\sqrt {9.2} + 2\sqrt {36.2} = \sqrt {100.2} \\
\Leftrightarrow \sqrt 2 x - 3.\sqrt {{3^2}.2} + 2\sqrt {{6^2}.2} = \sqrt {{{10}^2}.2} \\
\Leftrightarrow \sqrt 2 .x - 3.3\sqrt 2 + 2.6\sqrt 2 = 10\sqrt 2 \\
\Leftrightarrow \sqrt 2 .x - 9\sqrt 2 + 12\sqrt 2 = 10\sqrt 2 \\
\Leftrightarrow \sqrt 2 x = 10\sqrt 2 + 9\sqrt 2 - 12\sqrt 2 \\
\Leftrightarrow \sqrt 2 .x = 7\sqrt 2 \\
\Leftrightarrow x = 7\sqrt 2 :\sqrt 2 \\
\Leftrightarrow x = 7\\
d,\\
DKXD:\,\,\,{\left( {2x - 5} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {{{\left( {2x - 5} \right)}^2}} = 6\\
\Leftrightarrow \left| {2x - 5} \right| = 6\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 5 = 6\\
2x - 5 = - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 11\\
2x = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
e,\\
DKXD:\,\,\,9{x^2} - 12x + 4 \ge 0,\,\,\,\forall x\\
\sqrt {9{x^2} - 12x + 4} = \left| { - 10} \right|\\
\Leftrightarrow \sqrt {{{\left( {3x} \right)}^2} - 2.3x.2 + {2^2}} = 10\\
\Leftrightarrow \sqrt {{{\left( {3x - 2} \right)}^2}} = 10\\
\Leftrightarrow \left| {3x - 2} \right| = 10\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 2 = 10\\
3x - 2 = - 10
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = 12\\
3x = - 8
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = - \dfrac{8}{3}
\end{array} \right.
\end{array}\)
