Toán Lớp 9: Cho P= ($\frac{x+3\sqrt{x}+2 }{(x\sqrt{x}+2)(\sqrt{x}-1) }$ – $\frac{x+\sqrt{x} }{x-1}$) :($\frac{1}{\sqrt{x}+1}$ – $\frac{1}{\sqrt{x}-1 }$ )
a) Rút gọn P
b) Tìm x để $\frac{1}{p}$ -$\frac{\sqrt{x}+1 }{8}$ $\geq$ 1
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Toán Lớp 9: Cho P= ($\frac{x+3\sqrt{x}+2 }{(x\sqrt{x}+2)(\sqrt{x}-1) }$ – $\frac{x+\sqrt{x} }{x-1}$) :($\frac{1}{\sqrt{x}+1}$ – $\frac{1}{\sqrt{x}-1 }$ )
a) Rút gọn P
b) Tìm x để $\frac{1}{p}$ -$\frac{\sqrt{x}+1 }{8}$ $\geq$ 1
Comments ( 1 )
a)Dkxd:x \ge 0;x \ne 1\\
P = \left( {\dfrac{{x + 3\sqrt x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}} – \dfrac{{x + \sqrt x }}{{x – 1}}} \right)\\
:\left( {\dfrac{1}{{\sqrt x + 1}} – \dfrac{1}{{\sqrt x – 1}}} \right)\\
= \left( {\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}} – \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
:\dfrac{{\sqrt x – 1 – \sqrt x – 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}\\
= \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} – \dfrac{{\sqrt x }}{{\sqrt x – 1}}} \right).\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}{{ – 2}}\\
= \dfrac{1}{{\sqrt x – 1}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}{{ – 2}}\\
= \dfrac{{\sqrt x + 1}}{{ – 2}}\\
= – \dfrac{{\sqrt x + 1}}{2}\\
b)\dfrac{1}{P} – \dfrac{{\sqrt x + 1}}{8} \ge 1\\
\Leftrightarrow 1:\left( { – \dfrac{{\sqrt x + 1}}{2}} \right) – \dfrac{{\sqrt x + 1}}{8} – 1 \ge 0\\
\Leftrightarrow \dfrac{{ – 2}}{{\sqrt x + 1}} – \dfrac{{\sqrt x + 1}}{8} – 1 \ge 0\\
\Leftrightarrow \dfrac{{ – 16 – {{\left( {\sqrt x + 1} \right)}^2} – 8\left( {\sqrt x + 1} \right)}}{{8\left( {\sqrt x + 1} \right)}} \ge 0\\
\Leftrightarrow – 16 – x – 2\sqrt x – 1 – 8\sqrt x – 8 \ge 0\\
\Leftrightarrow – x – 26\sqrt x – 9 \ge 0\\
\Leftrightarrow x + 26\sqrt x + 9 \le 0\left( {ktm} \right)\\
Vậy\,x \in \emptyset
\end{array}$