Đáp án:
\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = \dfrac{3}{4}\\
x = \dfrac{1}{2}
\end{array} \right.\\
c,\\
x = 80
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,9{x^2} - 12x + 4 \ge 0,\,\,\,\forall x\\
\sqrt {9{x^2} - 12x + 4} = 1 - x\\
\Leftrightarrow \sqrt {{{\left( {3x} \right)}^2} - 2.3x.2 + {2^2}} = 1 - x\\
\Leftrightarrow \sqrt {{{\left( {3x - 2} \right)}^2}} = 1 - x\\
\Leftrightarrow \left| {3x - 2} \right| = 1 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
1 - x \ge 0\\
\left[ \begin{array}{l}
3x - 2 = 1 - x\\
3x - 2 = x - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
\left[ \begin{array}{l}
4x = 3\\
2x = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
\left[ \begin{array}{l}
x = \dfrac{3}{4}\\
x = \dfrac{1}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{4}\\
x = \dfrac{1}{2}
\end{array} \right.\\
c,\\
DKXD:\,\,\,x > - 1\\
\sqrt {4x + 4} + \sqrt {9x + 9} - \left( {x + 1} \right)\sqrt {\dfrac{4}{{x + 1}}} = 27\\
\Leftrightarrow \sqrt {4x + 4} + \sqrt {9\left( {x + 1} \right)} - \sqrt {{{\left( {x + 1} \right)}^2}.\dfrac{4}{{x + 1}}} = 27\\
\Leftrightarrow \sqrt {4x + 4} + \sqrt {{3^2}.\left( {x + 1} \right)} - \sqrt {4.\left( {x + 1} \right)} = 27\\
\Leftrightarrow \sqrt {4x + 4} + 3\sqrt {x + 1} - \sqrt {4x + 4} = 27\\
\Leftrightarrow 3\sqrt {x + 1} = 27\\
\Leftrightarrow \sqrt {x + 1} = 9\\
\Leftrightarrow x + 1 = {9^2}\\
\Leftrightarrow x + 1 = 81\\
\Leftrightarrow x = 80
\end{array}\)
