Đáp án:
\(\begin{array}{l}
a)B = \dfrac{1}{{\sqrt x + 1}}\\
b)\dfrac{1}{6}\\
c)\dfrac{9}{4} = x\\
d)\dfrac{1}{4} < x;x \ne 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)B = \dfrac{{4\left( {\sqrt x - 1} \right) - 2\left( {\sqrt x + 1} \right) - \sqrt x + 5}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{4\sqrt x - 4 - 2\sqrt x - 2 - \sqrt x + 5}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x + 1}}\\
b)Thay:x = 25\\
\to B = \dfrac{1}{{\sqrt {25} + 1}} = \dfrac{1}{6}\\
c)B = \dfrac{2}{5}\\
\to \dfrac{1}{{\sqrt x + 1}} = \dfrac{2}{5}\\
\to \dfrac{{5 - 2\sqrt x - 2}}{{5\left( {\sqrt x + 1} \right)}} = 0\\
\to 3 - 2\sqrt x = 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to \dfrac{9}{4} = x\\
d)B < \dfrac{2}{3}\\
\to \dfrac{1}{{\sqrt x + 1}} < \dfrac{2}{3}\\
\to \dfrac{{3 - 2\sqrt x - 2}}{{3\left( {\sqrt x + 1} \right)}} < 0\\
\to \dfrac{{1 - 2\sqrt x }}{{\sqrt x + 1}} < 0\\
\to 1 - 2\sqrt x < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to \dfrac{1}{4} < x;x \ne 1
\end{array}\)
