Đáp án:
B1: b. a=9
Giải thích các bước giải:
\(\begin{array}{l}
c.2\left| {x + 2} \right| = 8\\
\to \left| {x + 2} \right| = 4\\
\to \left[ \begin{array}{l}
x + 2 = 4\\
x + 2 = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 6
\end{array} \right.\\
d.DK:1 \ge x\\
\sqrt {1 - x} + 2\sqrt {1 - x} - \dfrac{1}{3}.4\sqrt {1 - x} + 5 = 0\\
\to \left( {1 + 2 - \dfrac{4}{3}} \right)\sqrt {1 - x} + 5 = 0\\
\to \dfrac{5}{3}\sqrt {1 - x} = - 5\\
\to \sqrt {1 - x} = - 3\left( {vô lý} \right)\\
\to x \in \emptyset \\
f.\sqrt[3]{{4x + 1}} = \sqrt[3]{{ - 7}}\\
\to 4x + 1 = - 7\\
\to 4x = - 8\\
\to x = - 2\\
B1:\\
a.P = \dfrac{{{{\left( {\sqrt a + 2} \right)}^2}}}{{\sqrt a + 2}} + \dfrac{{\left( {2 - \sqrt a } \right)\left( {\sqrt a + 2} \right)}}{{2 - \sqrt a }}\\
= \sqrt a + 2 + \sqrt a + 2 = 2\sqrt a + 4\\
b.P = a + 1\\
\to 2\sqrt a + 4 = a + 1\\
\to a - 2\sqrt a - 3 = 0\\
\to \left( {\sqrt a + 1} \right)\left( {\sqrt a - 3} \right) = 0\\
\to \sqrt a - 3 = 0\left( {do:\sqrt a + 1 > 0\forall a \ge 0} \right)\\
\to a = 9
\end{array}\)
