## Cho Q= √x+2/√x+3 – √x+1/√x-2 – 3(√x-1)/x-5√x+6 a, Rút gọn Q b, Tìm x để Q<1 c, Tìm x thuộc Z để Q thuộc R

Question

Cho Q= √x+2/√x+3 – √x+1/√x-2 – 3(√x-1)/x-5√x+6
a, Rút gọn Q
b, Tìm x để Q<1 c, Tìm x thuộc Z để Q thuộc R

in progress 0
15 phút 2021-09-08T05:02:17+00:00 1 Answers 0 views 0

$\begin{array}{l} DK:\left\{ \begin{array}{l} x > 0\\ x \ne 4\\ x \ne 9 \end{array} \right.\\ a,\\ Q = \frac{{\sqrt x + 2}}{{\sqrt x – 3}} – \frac{{\sqrt x + 1}}{{\sqrt x – 2}} – \frac{{3\left( {\sqrt x – 1} \right)}}{{x – 5\sqrt x + 6}}\\ \Leftrightarrow Q = \frac{{\sqrt x + 2}}{{\sqrt x – 3}} – \frac{{\sqrt x + 1}}{{\sqrt x – 2}} – \frac{{3\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}\\ \Leftrightarrow Q = \frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right) – \left( {\sqrt x + 1} \right)\left( {\sqrt x – 3} \right) – 3\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}\\ \Leftrightarrow Q = \frac{{x – 4 – \left( {x – 2\sqrt x – 3} \right) – 3\sqrt x + 3}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} = \frac{{ – \sqrt x + 2}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} = \frac{1}{{3 – \sqrt x }} \end{array}$
b,$\begin{array}{l} Q < 1 \Leftrightarrow \frac{1}{{3 – \sqrt x }} – 1 < 0\\ \Leftrightarrow \frac{{1 – 3 + \sqrt x }}{{3 – \sqrt x }} < 0 \Leftrightarrow \frac{{\sqrt x – 2}}{{\sqrt x – 3}} > 0 \Leftrightarrow \left[ \begin{array}{l} x > 9\\ x < 4 \end{array} \right. \end{array}$