P=( $\frac{2√x}{√x+3}$ + $\frac{√x}{√x-3}$ – $\frac{3x+3}{x-9}$ ) và Q= ( $\frac{√x+1}{√x-3}$; với x $\geq$ 0, x $\neq$ 9
1. Tính Q khi x=36
2. Tìm x để $\frac{P}{Q}$ < - $\frac{1}{2}$
Nhanh nhất đúng nhất auto 5 sao nhá thanks các bn
P=( $\frac{2√x}{√x+3}$ + $\frac{√x}{√x-3}$ – $\frac{3x+3}{x-9}$ ) và Q= ( $\frac{√x+1}{√x-3}$; với x $\geq$ 0, x $\neq$ 9 1. Tính Q khi x=36 2. Tì
By Reagan
Đáp án:
2) \(0 \le x < 9\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 36\\
Q = \dfrac{{\sqrt {36} + 1}}{{\sqrt {36} – 3}} = \dfrac{{6 + 1}}{{6 – 3}} = \dfrac{7}{3}\\
2)P = \dfrac{{2\sqrt x \left( {\sqrt x – 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) – 3x – 3}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x – 6\sqrt x + x + 3\sqrt x – 3x – 3}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ – 3\sqrt x – 3}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\
\dfrac{P}{Q} = \dfrac{{ – 3\sqrt x – 3}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\
= \dfrac{{ – 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\
= – \dfrac{3}{{\sqrt x + 3}}\\
\dfrac{P}{Q} < – \dfrac{1}{2}\\
\to – \dfrac{3}{{\sqrt x + 3}} < – \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 – \sqrt x – 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{3 – \sqrt x }}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to 3 – \sqrt x > 0\left( {do:\sqrt x + 3 > 0} \right)\\
\to 9 > x\\
\to 0 \le x < 9
\end{array}\)