Cho P= (x √x -3)/ (x- 2√x -3) – 2(√x -3 )/ (√x +1) = (√x +3) /( 3-√x)
a) Rút gọn P
b) Tính P khi x = 14 -6√5
c) Tính GTNN của P
Cho P= (x √x -3)/ (x- 2√x -3) – 2(√x -3 )/ (√x +1) = (√x +3) /( 3-√x)
a) Rút gọn P
b) Tính P khi x = 14 -6√5
c) Tính GTNN của P
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 9\\
P = \dfrac{{x\sqrt x – 3}}{{x – 2\sqrt x – 3}} – \dfrac{{2\left( {\sqrt x – 3} \right)}}{{\sqrt x + 1}} + \dfrac{{\sqrt x + 3}}{{3 – \sqrt x }}\\
= \dfrac{{x\sqrt x – 3 – 2\left( {\sqrt x – 3} \right)\left( {\sqrt x – 3} \right) – \left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x – 3 – 2\left( {x – 6\sqrt x + 9} \right) – \left( {x + 4\sqrt x + 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x – 3x + 8\sqrt x – 24}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x – 3} \right)\left( {x + 8} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 8}}{{\sqrt x + 1}}\\
b)x = 14 – 6\sqrt 5 \left( {tmdk} \right)\\
\Rightarrow x = {\left( {3 – \sqrt 5 } \right)^2}\\
\Rightarrow \sqrt x = 3 – \sqrt 5 \\
\Rightarrow P = \dfrac{{14 – 6\sqrt 5 + 8}}{{3 – \sqrt 5 + 1}}\\
= \dfrac{{22 – 6\sqrt 5 }}{{4 – \sqrt 5 }}\\
= \dfrac{{58 – 2\sqrt 5 }}{{11}}\\
c)P = \dfrac{{x + 8}}{{\sqrt x + 1}} = \dfrac{{x + \sqrt x – \sqrt x – 1 + 9}}{{\sqrt x + 1}}\\
= \sqrt x – 1 + \dfrac{9}{{\sqrt x + 1}}\\
= \sqrt x + 1 + \dfrac{9}{{\sqrt x + 1}} – 2\\
Theo\,Co – si:\\
\left( {\sqrt x + 1} \right) + \dfrac{9}{{\sqrt x + 1}} \ge 2\sqrt {\left( {\sqrt x + 1} \right).\dfrac{9}{{\sqrt x + 1}}} = 6\\
\Rightarrow P \ge 6 – 2 = 4\\
\Rightarrow GTNN:P = 4\\
Khi:\sqrt x + 1 = \dfrac{9}{{\sqrt x + 1}}\\
\Rightarrow \sqrt x + 1 = 3\\
\Rightarrow \sqrt x = 2\\
\Rightarrow x = 4\left( {tmdk} \right)
\end{array}$