Đáp án:
\(\begin{array}{l}
17,\\
A = \dfrac{{2x + 2\sqrt x + 2\sqrt 2 }}{x}\\
18,\\
a,\\
A = 9\\
b,\\
B = 1
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
17,\\
A = \left( {\dfrac{{x - 4}}{{\sqrt x - 2}} + \dfrac{{x + 2\sqrt 2 }}{{\sqrt x }}} \right):\sqrt x \\
= \left( {\dfrac{{{{\sqrt x }^2} - {2^2}}}{{\sqrt x - 2}} + \dfrac{{x + 2\sqrt 2 }}{{\sqrt x }}} \right):\sqrt x \\
= \left( {\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x - 2}} + \dfrac{{x + 2\sqrt 2 }}{{\sqrt x }}} \right):\sqrt x \\
= \left( {\sqrt x + 2 + \dfrac{{x + 2\sqrt 2 }}{{\sqrt x }}} \right):\sqrt x \\
= \dfrac{{\left( {\sqrt x + 2} \right).\sqrt x + x + 2\sqrt 2 }}{{\sqrt x }}.\dfrac{1}{{\sqrt x }}\\
= \dfrac{{{{\sqrt x }^2} + 2\sqrt x + x + 2\sqrt 2 }}{{{{\sqrt x }^2}}}\\
= \dfrac{{x + 2\sqrt x + x + 2\sqrt 2 }}{x}\\
= \dfrac{{2x + 2\sqrt x + 2\sqrt 2 }}{x}\\
18,\\
a,\\
A = \sqrt {16} + \sqrt {25} = \sqrt {{4^2}} + \sqrt {{5^2}} = \left| 4 \right| + \left| 5 \right| = 4 + 5 = 9\\
b,\\
B = \dfrac{{\left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt {x + 1} + 1} \right)}}{x}\\
= \dfrac{{{{\sqrt {x + 1} }^2} - {1^2}}}{x}\\
= \dfrac{{x + 1 - 1}}{x}\\
= \dfrac{x}{x}\\
= 1
\end{array}\)
