A=( 2x+1/x căn x -1 – 1/ căn x-1 ): ( 1- x-2/ x + căn x+1 )
a, rút gọn A
b, Tính A biết x = (2- căn 3) / 2
c tìm x thuộc z để A thuộc Z
A=( 2x+1/x căn x -1 – 1/ căn x-1 ): ( 1- x-2/ x + căn x+1 ) a, rút gọn A b, Tính A biết x = (2- căn 3) / 2 c tìm x thuộc z để A thuộc Z
By Julia
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 1\\
A = \left( {\dfrac{{2x + 1}}{{x\sqrt x – 1}} – \dfrac{1}{{\sqrt x – 1}}} \right):\left( {1 – \dfrac{{x – 2}}{{x + \sqrt x + 1}}} \right)\\
= \left( {\dfrac{{2x + 1}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}} – \dfrac{1}{{\sqrt x – 1}}} \right):\dfrac{{x + \sqrt x + 1 – x + 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{2x + 1 – \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 3}}\\
= \dfrac{{x – \sqrt x }}{{\sqrt x – 1}}.\dfrac{1}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
b)x = \dfrac{{2 – \sqrt 3 }}{2}\left( {tmdk} \right)\\
= \dfrac{{4 – 2\sqrt 3 }}{4}\\
= {\left( {\dfrac{{\sqrt 3 – 1}}{2}} \right)^2}\\
\Leftrightarrow \sqrt x = \dfrac{{\sqrt 3 – 1}}{2}\\
\Leftrightarrow A = \dfrac{{\sqrt x }}{{\sqrt x + 3}} = \dfrac{{\dfrac{{\sqrt 3 – 1}}{2}}}{{\dfrac{{\sqrt 3 – 1}}{2} + 3}}\\
= \dfrac{{\sqrt 3 – 1}}{{\sqrt 3 – 1 + 6}}\\
= \dfrac{{\left( {\sqrt 3 – 1} \right)\left( {5 – \sqrt 3 } \right)}}{{\left( {\sqrt 3 + 5} \right)\left( {5 – \sqrt 3 } \right)}}\\
= \dfrac{{6\sqrt 3 – 14}}{{25 – 3}}\\
= \dfrac{{3\sqrt 3 – 7}}{{11}}\\
c)A = \dfrac{{\sqrt x }}{{\sqrt x + 3}} = \dfrac{{\sqrt x + 3 – 3}}{{\sqrt x + 3}}\\
= 1 – \dfrac{3}{{\sqrt x + 3}}\\
A \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 = 3\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\left( {tmdk} \right)\\
Vậy\,x = 0
\end{array}$