giúp em bài cuối với Question giúp em bài cuối với in progress 0 Môn Toán Philomena 5 years 2020-11-10T13:23:01+00:00 2020-11-10T13:23:01+00:00 1 Answers 64 views 0
Answers ( )
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
Q = \left( {\dfrac{{x – \sqrt x }}{{\sqrt x – 1}} – \dfrac{{\sqrt x + 1}}{{x + \sqrt x }}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \left( {\dfrac{{\sqrt x \left( {\sqrt x – 1} \right)}}{{\sqrt x – 1}} – \dfrac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \left( {\sqrt x – \dfrac{1}{{\sqrt x }}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{{{\sqrt x }^2} – 1}}{{\sqrt x }}:\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
= \sqrt x – 1\\
b,\\
x = 12 + 8\sqrt 2 = 8 + 2.2\sqrt 2 .2 + 4 = {\left( {2\sqrt 2 } \right)^2} + 2.2\sqrt 2 .2 + {2^2} = {\left( {2\sqrt 2 + 2} \right)^2}\\
\Rightarrow \sqrt x = 2\sqrt 2 + 2\\
\Rightarrow Q = \left( {2\sqrt 2 + 2} \right) – 1 = 2\sqrt 2 + 1
\end{array}\)