Đáp án:
B2:
b) a=9
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)A = \dfrac{{x\sqrt x - 2x + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {x - 2\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x \left( {\sqrt x - 1} \right)}} = \sqrt x - 1\\
b)Thay:x = 3 + 2\sqrt 2 = {\left( {\sqrt 2 + 1} \right)^2}\\
\to A = \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - 1 = \sqrt 2 + 1 - 1 = \sqrt 2 \\
B2:\\
a)P = \dfrac{{{{\left( {\sqrt a + 2} \right)}^2}}}{{\sqrt a + 2}} + \dfrac{{\left( {2 - \sqrt a } \right)\left( {\sqrt a + 2} \right)}}{{2 - \sqrt a }}\\
= \sqrt a + 2 + \sqrt a + 2 = 2\left( {\sqrt a + 2} \right)\\
b)P = a + 1\\
\to 2\sqrt a + 4 = a + 1\\
\to a - 2\sqrt a - 3 = 0\\
\to \left( {\sqrt a - 3} \right)\left( {\sqrt a + 1} \right) = 0\\
\to \sqrt a - 3 = 0\left( {do:\sqrt a + 1 > 0\forall a \ge 0} \right)\\
\to a = 9\\
B3:\\
a)DK:x \ge 0;x \ne 1\\
b)A = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}} + \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \sqrt x - 1 + \sqrt x = 2\sqrt x - 1\\
c)A < - 1\\
\to 2\sqrt x - 1 < - 1\\
\to 2\sqrt x < 0\left( {vô lý} \right)\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to x \in \emptyset
\end{array}\)
