Đáp án:
\(\begin{array}{l}
a)\dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
b)\dfrac{3}{2}\\
c)dpcm
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x > 0;x \ne \left\{ {\dfrac{1}{4};1} \right\}\\
P = \left( {1 - \sqrt x } \right).\left[ {\dfrac{{\sqrt x - 1 + \sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}} \right].\dfrac{{x + \sqrt x - 2\sqrt x + 1}}{{2\sqrt x - 1}}\\
= \left( {1 - \sqrt x } \right).\dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}.\dfrac{{x - \sqrt x + 1}}{{2\sqrt x - 1}}\\
= \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
b)Thay:x = \left( {8 + 2\sqrt {15} } \right)\left( {\sqrt 5 - \sqrt 3 } \right).\sqrt 2 .\sqrt {4 - \sqrt {15} } \\
= \left( {8 + 2\sqrt {15} } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {8 - 2\sqrt {15} } \\
= \left( {8 + 2\sqrt {15} } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {5 - 2\sqrt 5 .\sqrt 3 + 3} \\
= \left( {8 + 2\sqrt {15} } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \\
= \left( {8 + 2\sqrt {15} } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\\
= \left( {8 + 2\sqrt {15} } \right)\left( {8 - 2\sqrt {15} } \right)\\
= 64 - 4.15 = 4\\
\to P = \dfrac{{4 - \sqrt 4 + 1}}{{\sqrt 4 }} = \dfrac{3}{2}\\
c)P > 1\\
\to \dfrac{{x - \sqrt x + 1}}{{\sqrt x }} > 1\\
\to \dfrac{{x - \sqrt x + 1 - \sqrt x }}{{\sqrt x }} > 0\\
\to \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x }} > 0\\
\to \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} > 0\\
Do:x > 0;x \ne 1 \to \left\{ \begin{array}{l}
\sqrt x > 0\\
{\left( {\sqrt x - 1} \right)^2} > 0
\end{array} \right.\\
\to dpcm
\end{array}\)
