Đáp án:
\(\begin{array}{l}
a,\\
A = 1\\
B = 2\sqrt x - 1\\
b,\\
x > 1
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \sqrt {50} - 3\sqrt 8 + \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} \\
= \sqrt {25.2} - 3\sqrt {4.2} + \left| {\sqrt 2 + 1} \right|\\
= \sqrt {{5^2}.2} - 3\sqrt {{2^2}.2} + \left( {\sqrt 2 + 1} \right)\\
= 5\sqrt 2 - 3.2\sqrt 2 + \sqrt 2 + 1\\
= 5\sqrt 2 - 6\sqrt 2 + \sqrt 2 + 1\\
= 1\\
B = \dfrac{{x\sqrt x - \sqrt x }}{{x - 1}} + \dfrac{{x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x \left( {x - 1} \right)}}{{x - 1}} + \dfrac{{{{\sqrt x }^2} - {1^2}}}{{\sqrt x + 1}}\\
= \sqrt x + \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \sqrt x + \left( {\sqrt x - 1} \right)\\
= 2\sqrt x - 1\\
b,\\
A \le B\\
\Leftrightarrow 1 \le 2\sqrt x - 1\\
\Leftrightarrow 2 \le 2\sqrt x \\
\Leftrightarrow \sqrt x \ge 1\\
\Leftrightarrow x \ge 1\\
x \ge 0,x \ne 1 \Rightarrow x > 1
\end{array}\)
