Cho A= √x+1/ √x + 2/ 1-√x + 3√x+1/ x- √x với x>0 , x ≠ 1 a. tính giá trị biểu thức khi x=4+2√3 b. tìm x lớn nhất để A>1/2 c. tìm x để A nhận giá trị nguyên
1 câu trả lời
Đáp án:
$\begin{array}{l}
A = \dfrac{{\sqrt x + 1}}{{\sqrt x }} + \dfrac{2}{{1 - \sqrt x }} + \dfrac{{3\sqrt x + 1}}{{x - \sqrt x }}\left( {x > 0;x \ne 1} \right)\\
a)A = \dfrac{{\sqrt x + 1}}{{\sqrt x }} + \dfrac{2}{{1 - \sqrt x }} + \dfrac{{3\sqrt x + 1}}{{x - \sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - 2\sqrt x + 3\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 1 - 2\sqrt x + 3\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
x = 4 + 2\sqrt 3 \left( {tmdk} \right)\\
= {\left( {\sqrt 3 + 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 + 1\\
\Leftrightarrow A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt 3 + 1 + 1}}{{\sqrt 3 + 1 - 1}} = \dfrac{{\sqrt 3 + 2}}{{\sqrt 3 }} = \dfrac{{3 + 2\sqrt 3 }}{3}\\
b)A > \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{1}{2} > 0\\
\Leftrightarrow \dfrac{{2\sqrt x + 2 - \sqrt x + 1}}{{2\left( {\sqrt x - 1} \right)}} > 0\\
\Leftrightarrow \dfrac{{\sqrt x + 3}}{{2\left( {\sqrt x - 1} \right)}} > 0\\
\Leftrightarrow \sqrt x - 1 > 0\\
\Leftrightarrow \sqrt x > 1\\
\Leftrightarrow x > 1\\
Vậy\,x > 1\\
c)A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\sqrt x - 1 + 2}}{{\sqrt x - 1}}\\
= 1 + \dfrac{2}{{\sqrt x - 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \left( {\sqrt x - 1} \right) \in \left\{ {1;2} \right\}\left( {do:\sqrt x - 1 > - 1} \right)\\
\Leftrightarrow \sqrt x \in \left\{ {2;3} \right\}\\
\Leftrightarrow x \in \left\{ {4;9} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {4;9} \right\}
\end{array}$