Đáp án:
\(\begin{array}{l}
8A,\\
a,\\
\left[ \begin{array}{l}
x = - 1\\
x = 3
\end{array} \right.\\
b,\\
x = - 2\\
8B,\\
a,\\
\left[ \begin{array}{l}
x = 2\\
x = \dfrac{9}{4}
\end{array} \right.\\
b,\\
x = 3
\end{array}\)
\(\begin{array}{l}
9A,\\
y = 28\\
9B,\\
y = 9
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
8A,\\
a,\\
DKXD:\,\,\,x \ge - 1\\
x + 1 - 2\sqrt {x + 1} = 0\\
\Leftrightarrow {\sqrt {x + 1} ^2} - 2\sqrt {x + 1} = 0\\
\Leftrightarrow \sqrt {x + 1} .\left( {\sqrt {x + 1} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 1} = 0\\
\sqrt {x + 1} - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 1} = 0\\
\sqrt {x + 1} = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x + 1 = 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 3
\end{array} \right.\\
b,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
{x^2} - 2x \ge 0\\
2 - 3x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x\left( {x - 2} \right) \ge 0\\
3x \le 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x - 2 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x - 2 \le 0
\end{array} \right.
\end{array} \right.\\
x \le \dfrac{2}{3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x \ge 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x \le 2
\end{array} \right.
\end{array} \right.\\
x \le \dfrac{2}{3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
x \le \dfrac{2}{3}
\end{array} \right. \Leftrightarrow x \le 0\\
\sqrt {{x^2} - 2x} = \sqrt {2 - 3x} \\
\Leftrightarrow {\sqrt {{x^2} - 2x} ^2} = {\sqrt {2 - 3x} ^2}\\
\Leftrightarrow {x^2} - 2x = 2 - 3x\\
\Leftrightarrow {x^2} - 2x - 2 + 3x = 0\\
\Leftrightarrow {x^2} + x - 2 = 0\\
\Leftrightarrow \left( {{x^2} - x} \right) + \left( {2x - 2} \right) = 0\\
\Leftrightarrow x\left( {x - 1} \right) + 2.\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.\\
x \le 0 \Rightarrow x = - 2\\
8B,\\
a,\\
DKXD:\,\,\,x \ge 2\\
2x - 4 - \sqrt {x - 2} = 0\\
\Leftrightarrow 2.\left( {x - 2} \right) - \sqrt {x - 2} = 0\\
\Leftrightarrow 2{\sqrt {x - 2} ^2} - \sqrt {x - 2} = 0\\
\Leftrightarrow \sqrt {x - 2} .\left( {2\sqrt {x - 2} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 2} = 0\\
2\sqrt {x - 2} - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 2} = 0\\
\sqrt {x - 2} = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x - 2 = \dfrac{1}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{9}{4}
\end{array} \right.\\
b,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x - 3 \ge 0\\
{x^2} - 9 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
{x^2} \ge 9
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.
\end{array} \right. \Leftrightarrow x \ge 3\\
\sqrt {x - 3} - 2\sqrt {{x^2} - 9} = 0\\
\Leftrightarrow \sqrt {x - 3} - 2.\sqrt {{x^2} - {3^2}} = 0\\
\Leftrightarrow \sqrt {x - 3} - 2.\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} = 0\\
\Leftrightarrow \sqrt {x - 3} \left( {1 - 2\sqrt {x + 3} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 3} = 0\\
1 - 2\sqrt {x + 3} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 3} = 0\\
\sqrt {x + 3} = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x + 3 = \dfrac{1}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{{11}}{4}
\end{array} \right.\\
x \ge 3 \Rightarrow x = 3
\end{array}\)
\(\begin{array}{l}
9A,\\
DKXD:\,\,\,y \ge 3\\
2\sqrt {9y - 27} - \dfrac{1}{5}\sqrt {25y - 75} - \dfrac{1}{7}\sqrt {49y - 147} = 20\\
\Leftrightarrow 2.\sqrt {9\left( {y - 3} \right)} - \dfrac{1}{5}\sqrt {25.\left( {y - 3} \right)} - \dfrac{1}{7}\sqrt {49.\left( {y - 3} \right)} = 20\\
\Leftrightarrow 2\sqrt {{3^2}\left( {y - 3} \right)} - \dfrac{1}{5}\sqrt {{5^2}.\left( {y - 3} \right)} - \dfrac{1}{7}\sqrt {{7^2}.\left( {y - 3} \right)} = 20\\
\Leftrightarrow 2.3\sqrt {y - 3} - \dfrac{1}{5}.5\sqrt {y - 3} - \dfrac{1}{7}.7\sqrt {y - 3} = 20\\
\Leftrightarrow 6\sqrt {y - 3} - \sqrt {y - 3} - \sqrt {y - 3} = 20\\
\Leftrightarrow 4\sqrt {y - 3} = 20\\
\Leftrightarrow \sqrt {y - 3} = 5\\
\Leftrightarrow y - 3 = {5^2}\\
\Leftrightarrow y - 3 = 25\\
\Leftrightarrow y = 25 + 3\\
\Leftrightarrow y = 28\\
9B,\\
DKXD:\,\,\,y \ge 5\\
\sqrt {4y - 20} + \sqrt {y - 5} - \dfrac{1}{3}\sqrt {9y - 45} = 4\\
\Leftrightarrow \sqrt {4\left( {y - 5} \right)} + \sqrt {y - 5} - \dfrac{1}{3}\sqrt {9.\left( {y - 5} \right)} = 4\\
\Leftrightarrow \sqrt {{2^2}.\left( {y - 5} \right)} + \sqrt {y - 5} - \dfrac{1}{3}.\sqrt {{3^2}.\left( {y - 5} \right)} = 4\\
\Leftrightarrow 2\sqrt {y - 5} + \sqrt {y - 5} - \dfrac{1}{3}.3\sqrt {y - 5} = 4\\
\Leftrightarrow 2\sqrt {y - 5} + \sqrt {y - 5} - \sqrt {y - 5} = 4\\
\Leftrightarrow 2\sqrt {y - 5} = 4\\
\Leftrightarrow \sqrt {y - 5} = 2\\
\Leftrightarrow y - 5 = {2^2}\\
\Leftrightarrow y - 5 = 4\\
\Leftrightarrow y = 4 + 5\\
\Leftrightarrow y = 9
\end{array}\)
