Đáp án:
\(\begin{array}{l}
B1.26:\\
a)B = \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)x = 9\\
B1.27:\\
1)A = 14\sqrt 3 \\
2)a)B = \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
b)x = 16
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1.26:\\
a)B = \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)\dfrac{B}{A} = \dfrac{4}{3}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x }}:\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{4}{3}\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x }} = \dfrac{4}{3}\\
\to 3\sqrt x + 3 = 4\sqrt x \\
\to \sqrt x = 3\\
\to x = 9\\
B1.27:\\
1)A = \left( {2\sqrt 3 - 2\sqrt 3 + 5\sqrt 2 - \dfrac{3}{4}.2\sqrt 2 } \right).2\sqrt 6 \\
= \dfrac{7}{2}\sqrt 2 .2\sqrt 6 = 14\sqrt 3 \\
2)a)B = \dfrac{{x - 1 - 4\sqrt x + \sqrt x + 1}}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
b)B = \dfrac{1}{2} \to \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} = \dfrac{1}{2}\\
\to 2\sqrt x - 6 = \sqrt x - 2\\
\to \sqrt x = 4\\
\to x = 16
\end{array}\)
