Đáp án:
$\begin{array}{l}
B7)\\
Dkxd:a \ge 0;a\# 4\\
R = \left( {\dfrac{{a - \sqrt a + 7}}{{a - 4}} + \dfrac{1}{{\sqrt a - 2}}} \right):\left( {\dfrac{{\sqrt a + 2}}{{\sqrt a - 2}} - \dfrac{{\sqrt a - 2}}{{\sqrt a + 2}} - \dfrac{{2\sqrt a }}{{a - 4}}} \right)\\
= \dfrac{{a - \sqrt a + 7 + \sqrt a + 2}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}:\dfrac{{{{\left( {\sqrt a + 2} \right)}^2} - {{\left( {\sqrt a - 2} \right)}^2} - 2\sqrt a }}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}\\
= \dfrac{{a + 9}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}.\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}{{4\sqrt a - 2\sqrt a }}\\
= \dfrac{{a + 9}}{{2\sqrt a }}\\
B8)\\
a)Dkxd:x > 0;x\# 4;x\# 9\\
K = \left( {\dfrac{{2 + \sqrt x }}{{2 - \sqrt x }} - \dfrac{{2 - \sqrt x }}{{2 + \sqrt x }} - \dfrac{{4x}}{{x - 4}}} \right):\dfrac{{\sqrt x - 3}}{{2\sqrt x - x}}\\
= \dfrac{{{{\left( {2 + \sqrt x } \right)}^2} - {{\left( {2 - \sqrt x } \right)}^2} + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x - 3}}\\
= \dfrac{{8\sqrt x + 4x}}{{2 + \sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
= \dfrac{{4\sqrt x \left( {2 + \sqrt x } \right)}}{{2 + \sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
= \dfrac{{4x}}{{\sqrt x - 3}}\\
b)K > 0\\
\Leftrightarrow \dfrac{{4x}}{{\sqrt x - 3}} > 0\\
\Leftrightarrow \sqrt x - 3 > 0\left( {do:x > 0} \right)\\
\Leftrightarrow \sqrt x > 3\\
\Leftrightarrow x > 9\\
Vay\,x > 9\\
c)K = 1\\
\Leftrightarrow \dfrac{{4x}}{{\sqrt x - 3}} = 1\\
\Leftrightarrow 4x = \sqrt x - 3\\
\Leftrightarrow 4x - \sqrt x + 3 = 0\\
\Leftrightarrow {\left( {2\sqrt x } \right)^2} - 2.2\sqrt x .\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{47}}{{16}} = 0\\
\Leftrightarrow {\left( {2\sqrt x - \dfrac{1}{4}} \right)^2} + \dfrac{{47}}{{16}} = 0\left( {vn} \right)
\end{array}$
Vậy ko có x thỏa mãn để K=1
