P=( $\frac{√x-2}{√x-1}$ – $\frac{ √x+2}{x+2√x+1}$ ). $\frac{(1-x)^2}{2}$ .Với > 0;x $\neq$ 1
a)Rút gọn P
b)Tính trị của P khi x= 7-4 √3
c) Tìm x để P có GTLN
P=( $\frac{√x-2}{√x-1}$ – $\frac{ √x+2}{x+2√x+1}$ ). $\frac{(1-x)^2}{2}$ .Với > 0;x $\neq$ 1
a)Rút gọn P
b)Tính trị của P khi x= 7-4 √3
c) Tìm x để P có GTLN
Đáp án:
$\begin{array}{l}
a)P = \left( {\frac{{\sqrt x – 2}}{{\sqrt x – 1}} – \frac{{\sqrt x + 2}}{{x – 2\sqrt x + 1}}} \right).\frac{{{{\left( {1 – \sqrt x } \right)}^2}}}{2}\left( {x > 0;x \ne 1} \right)\\
= \frac{{\left( {\sqrt x – 2} \right).\left( {\sqrt x – 1} \right) – \left( {\sqrt x + 2} \right)}}{{{{\left( {\sqrt x – 1} \right)}^2}}}.\frac{{{{\left( {\sqrt x – 1} \right)}^2}}}{2}\\
= \frac{{x – 3\sqrt x + 2 – \sqrt x – 2}}{2}\\
= \frac{{x – 4\sqrt x }}{2}\\
b)x = 7 – 4\sqrt 3 \left( {tmdk} \right)\\
\Rightarrow x = 4 – 2.2\sqrt 3 + 3 = {\left( {2 – \sqrt 3 } \right)^2}\\
\Rightarrow \sqrt x = 2 – \sqrt 3 \\
\Rightarrow P = \frac{{x – 4\sqrt x }}{2}\\
= \frac{{7 – 4\sqrt 3 – 4\left( {2 – \sqrt 3 } \right)}}{2}\\
= \frac{{7 – 4\sqrt 3 – 8 + 4\sqrt 3 }}{2}\\
= \frac{{ – 1}}{2}\\
c)P = \frac{{x – 4\sqrt x }}{2} = \frac{{\left( {x – 4\sqrt x + 4} \right) – 4}}{2} = \frac{{{{\left( {\sqrt x – 2} \right)}^2}}}{2} – 2\\
Do:{\left( {\sqrt x – 2} \right)^2} \ge 0\forall x\\
\Rightarrow P \ge – 2\forall x\\
Dau\, = \,xay\,ra \Leftrightarrow \sqrt x = 2\\
\Rightarrow x = 4\left( {tmdk} \right)
\end{array}$