Toán Lớp 9: P =√x/√x-1+3/√x+1-6√x-4/x-1
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Toán Lớp 9: P =√x/√x-1+3/√x+1-6√x-4/x-1
Tìm giá trị nhỏ nhất của P
Comments ( 2 )
Dkxd:x \ge 0;x \ne 1\\
P = \dfrac{{\sqrt x }}{{\sqrt x – 1}} + \dfrac{3}{{\sqrt x + 1}} – \dfrac{{6\sqrt x – 4}}{{x – 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + 3\left( {\sqrt x – 1} \right) – 6\sqrt x + 4}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + \sqrt x + 3\sqrt x – 3 – 6\sqrt x + 4}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x – 2\sqrt x + 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x – 1}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1 – 2}}{{\sqrt x + 1}}\\
= 1 – \dfrac{2}{{\sqrt x + 1}}\\
Do:\sqrt x + 1 \ge 1\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 1}} \le 1\\
\Leftrightarrow \dfrac{2}{{\sqrt x + 1}} \le 2\\
\Leftrightarrow – \dfrac{2}{{\sqrt x + 1}} \ge – 2\\
\Leftrightarrow 1 – \dfrac{2}{{\sqrt x + 1}} \ge – 1\\
\Leftrightarrow P \ge – 1\\
\Leftrightarrow GTNN:P = – 1\,khi:x = 0
\end{array}$