Đáp án:

\(\begin{array}{l}
B7:\\
a)\dfrac{2}{{x - 1}}\\
b)2\\
c)3 > x > 1\\
d)\left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
B8:\\
a)\dfrac{{\sqrt a  - 1}}{{\sqrt a }}\\
b)a = \dfrac{1}{4}\\
c)M < 1\\
d)0 < a < 1
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B7:\\
a)DK:x > 0;x \ne 1\\
C = \dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right) - \left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  + 1} \right)}^2}\left( {\sqrt x  - 1} \right)}}.\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \dfrac{{x + \sqrt x  - 2 - x + \sqrt x  + 2}}{{{{\left( {\sqrt x  + 1} \right)}^2}\left( {\sqrt x  - 1} \right)}}.\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \dfrac{{2\sqrt x }}{{{{\left( {\sqrt x  + 1} \right)}^2}\left( {\sqrt x  - 1} \right)}}.\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \dfrac{2}{{x - 1}}\\
b)\left| {2x - 3} \right| = 1\\
 \to \left[ \begin{array}{l}
2x - 3 = 1\\
2x - 3 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x = 1\left( l \right)
\end{array} \right.\\
Thay:x = 2\\
 \to C = \dfrac{2}{{2 - 1}} = 2\\
c)C > 1\\
 \to \dfrac{2}{{x - 1}} > 1\\
 \to \dfrac{{2 - x + 1}}{{x - 1}} > 0\\
 \to \dfrac{{3 - x}}{{x - 1}} > 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3 - x > 0\\
x - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3 - x < 0\\
x - 1 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
3 > x > 1\\
\left\{ \begin{array}{l}
x > 3\\
x < 1
\end{array} \right.\left( l \right)
\end{array} \right.\\
d)C \in Z \to \dfrac{2}{{x - 1}} \in Z\\
 \to x - 1 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
x - 1 = 2\\
x - 1 =  - 2\\
x - 1 =  - 1\\
x - 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x =  - 1\left( l \right)\\
x = 0\left( l \right)\\
x = 2
\end{array} \right.\\
B8:\\
a)DK:a > 0;a \ne 1\\
M = \dfrac{{1 + \sqrt a }}{{\sqrt a \left( {\sqrt a  - 1} \right)}}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{\sqrt a  + 1}}\\
 = \dfrac{{\sqrt a  - 1}}{{\sqrt a }}\\
b)M =  - 1\\
 \to \dfrac{{\sqrt a  - 1}}{{\sqrt a }} =  - 1\\
 \to \sqrt a  - 1 =  - \sqrt a \\
 \to 2\sqrt a  = 1\\
 \to \sqrt a  = \dfrac{1}{2}\\
 \to a = \dfrac{1}{4}\\
c)M - 1 = \dfrac{{\sqrt a  - 1}}{{\sqrt a }} - 1\\
 = \dfrac{{\sqrt a  - 1 - \sqrt a }}{{\sqrt a }} =  - \dfrac{1}{{\sqrt a }}\\
Do:\sqrt a  > 0\forall a > 0\\
 \to  - \dfrac{1}{{\sqrt a }} < 0\\
 \to M - 1 < 0\\
 \to M < 1\\
d)M < 0\\
 \to \dfrac{{\sqrt a  - 1}}{{\sqrt a }} < 0\\
 \to \sqrt a  - 1 < 0\left( {do:\sqrt a  > 0\forall a > 0} \right)\\
 \to 0 < a < 1
\end{array}\)