Đáp án:
B13
b) \(a = \dfrac{1}{{16}}\)
Giải thích các bước giải:
\(\begin{array}{l}
C13:\\
a)P = \left[ {\dfrac{{1 + \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right].\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}}\\
= \dfrac{1}{{\sqrt a }}\\
b)P = 4\\
\to \dfrac{1}{{\sqrt a }} = 4\\
\to \sqrt a = \dfrac{1}{4}\\
\to a = \dfrac{1}{{16}}\\
B14:\\
\left\{ \begin{array}{l}
2x - y = 5\\
x + y = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3x = 9\\
x + y = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 3\\
y = 1
\end{array} \right.
\end{array}\)
