cho A = √x / √x-1 + 3/ √x+1 - 6 √x -4 /x-1 -1 a, rút gon A b,Tìm x để A = -2 c,Tìm x nguyên để A cũng là số nguyên
1 câu trả lời
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{\sqrt x }}{{\sqrt x - 1}} + \dfrac{3}{{\sqrt x + 1}} - \dfrac{{6\sqrt x - 4}}{{x - 1}} - 1\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + 3\left( {\sqrt x - 1} \right) - 6\sqrt x + 4 - x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + \sqrt x + 3\sqrt x - 3 - 6\sqrt x + 4 - x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 2\sqrt x + 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 2}}{{\sqrt x + 1}}\\
b)A = - 2\\
\Leftrightarrow \dfrac{{ - 2}}{{\sqrt x + 1}} = - 2\\
\Leftrightarrow \sqrt x + 1 = 1\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\left( {tmdk} \right)\\
Vay\,x = 0\\
c)A = \dfrac{{ - 2}}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \left( {\sqrt x + 1} \right) \in \left\{ {1;2} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {0;1} \right\}\\
\Leftrightarrow x \in \left\{ {0;1} \right\}\\
Do:x \ne 1\\
\Leftrightarrow x = 0\\
Vậy\,x = 0
\end{array}$