Đáp án:
\(\begin{array}{l}
5,\\
\left[ \begin{array}{l}
x = 1\\
x = - \dfrac{2}{3}
\end{array} \right.\\
6,\\
\left[ \begin{array}{l}
x = 1\\
x = \dfrac{5}{3}
\end{array} \right.\\
7,\\
x = 2\\
8,\\
x = \dfrac{1}{2}\\
9,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
10,\\
x = \dfrac{6}{5}\,
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
5,\\
DKXD:\,\,36{x^2} - 12x + 1 \ge 0\\
\sqrt {36{x^2} - 12x + 1} = 5\\
\Leftrightarrow \sqrt {{{\left( {6x} \right)}^2} - 2.6x.1 + {1^2}} = 5\\
\Leftrightarrow \sqrt {{{\left( {6x - 1} \right)}^2}} = 5\\
\Leftrightarrow \left| {6x - 1} \right| = 5\\
\Leftrightarrow \left[ \begin{array}{l}
6x - 1 = 5\\
6x - 1 = - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
6x = 6\\
6x = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{2}{3}
\end{array} \right.\\
6,\\
DKXD:\,\,\left\{ \begin{array}{l}
{x^2} - 4x + 4 \ge 0\\
4{x^2} - 12x + 9 \ge 0
\end{array} \right.\\
\sqrt {{x^2} - 4x + 4} = \sqrt {4{x^2} - 12x + 9} \\
\Leftrightarrow \sqrt {{x^2} - 2.x.2 + {2^2}} = \sqrt {{{\left( {2x} \right)}^2} - 2.2x.3 + {3^2}} \\
\Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = \sqrt {{{\left( {2x - 3} \right)}^2}} \\
\Leftrightarrow \left| {x - 2} \right| = \left| {2x - 3} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 2x - 3\\
x - 2 = - \left( {2x - 3} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 - \left( {2x - 3} \right) = 0\\
x - 2 + \left( {2x - 3} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 - 2x + 3 = 0\\
x - 2 + 2x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- x + 1 = 0\\
3x - 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{5}{3}
\end{array} \right.\\
7,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2{x^2} - 3 \ge 0\\
4x - 3 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} \ge \dfrac{3}{2}\\
x \ge \dfrac{3}{4}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \sqrt {\dfrac{3}{2}} \\
x \le - \sqrt {\dfrac{3}{2}}
\end{array} \right.\\
x \ge \dfrac{3}{4}
\end{array} \right. \Leftrightarrow x \ge \sqrt {\dfrac{3}{2}} \Leftrightarrow x \ge \dfrac{{\sqrt 6 }}{2}\\
\sqrt {2{x^2} - 3} = \sqrt {4x - 3} \\
\Leftrightarrow {\sqrt {2{x^2} - 3} ^2} = {\sqrt {4x - 3} ^2}\\
\Leftrightarrow 2{x^2} - 3 = 4x - 3\\
\Leftrightarrow 2{x^2} - 4x = 0\\
\Leftrightarrow 2x.\left( {x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
x \ge \dfrac{{\sqrt 6 }}{4} \Rightarrow x = 2\\
8,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
\dfrac{{2x - 3}}{{x - 1}} \ge 0\\
x - 1 \ne 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - 3 \ge 0\\
x - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 3 \le 0\\
x - 1 < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x > 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le \dfrac{3}{2}\\
x < 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x < 1
\end{array} \right.\\
\sqrt {\dfrac{{2x - 3}}{{x - 1}}} = 2\\
\Leftrightarrow \dfrac{{2x - 3}}{{x - 1}} = {2^2}\\
\Leftrightarrow \dfrac{{2x - 3}}{{x - 1}} = 4\\
\Leftrightarrow 2x - 3 = 4.\left( {x - 1} \right)\\
\Leftrightarrow 2x - 3 = 4x - 4\\
\Leftrightarrow 2x - 3 - 4x + 4 = 0\\
\Leftrightarrow - 2x + 1 = 0\\
\Leftrightarrow x = \dfrac{1}{2}\,\,\,\,\left( {t/m} \right)\\
9,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 1 \ge 0\\
x - 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \ge 1
\end{array} \right. \Leftrightarrow x \ge 1\\
\sqrt {2x - 1} = \sqrt {x - 1} \\
\Leftrightarrow {\sqrt {2x - 1} ^2} = {\sqrt {x - 1} ^2}\\
\Leftrightarrow 2x - 1 = x - 1\\
\Leftrightarrow 2x = x\\
\Leftrightarrow x = 0\\
x \ge 1 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
10,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
\dfrac{{4x + 3}}{{x + 1}} \ge 0\\
x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4x + 3 \ge 0\\
x + 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
4x + 3 \le 0\\
x + 1 < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - \dfrac{3}{4}\\
x > - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - \dfrac{3}{4}\\
x < - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge - \dfrac{3}{4}\\
x < - 1
\end{array} \right.\\
\sqrt {\dfrac{{4x + 3}}{{x + 1}}} = 3\\
\Leftrightarrow \dfrac{{4x + 3}}{{x + 1}} = {3^2}\\
\Leftrightarrow \dfrac{{4x + 3}}{{x + 1}} = 9\\
\Leftrightarrow 4x + 3 = 9.\left( {x + 1} \right)\\
\Leftrightarrow 4x + 3 = 9x + 9\\
\Leftrightarrow 9x + 9 - 4x - 3 = 0\\
\Leftrightarrow 5x - 6 = 0\\
\Leftrightarrow x = \dfrac{6}{5}\,\,\,\,\left( {t/m} \right)
\end{array}\)
