Toán Lớp 10: Tìm Min của biểu thức sau:
a) $A= \dfrac{1}{5-2x} + \dfrac{1}{3x-2} ( \dfrac{2}{3}
Toán Lớp 10: Tìm Min của biểu thức sau: a) $A= \dfrac{1}{5-2x} + \dfrac{1}{3x-2} ( \dfrac{2}{3}
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\dfrac{2}{3} < x < \dfrac{5}{2} \Rightarrow \left\{ \begin{array}{l}
3x – 2 > 0\\
5 – 2x > 0
\end{array} \right.\\
A = \dfrac{1}{{3x – 2}} + \dfrac{1}{{5 – 2x}}\\
= \dfrac{2}{{6x – 4}} + \dfrac{3}{{15 – 6x}}\\
= \dfrac{{{{\sqrt 2 }^2}}}{{6x – 4}} + \dfrac{{{{\sqrt 3 }^2}}}{{15 – 6x}}\\
\ge \dfrac{{{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2}}}{{6x – 4 + 15 – 6x}} = \dfrac{{{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2}}}{{11}}\left( {B – C – S} \right)\\
\Rightarrow \min A = \dfrac{{{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2}}}{{11}}\\
‘ = ‘ \Leftrightarrow \dfrac{{\sqrt 2 }}{{6x – 4}} = \dfrac{{\sqrt 3 }}{{15 – 6x}} \Leftrightarrow 15\sqrt 2 – 6\sqrt 2 x = 6\sqrt 3 x – 4\sqrt 3 \\
\Leftrightarrow 6\left( {\sqrt 2 + \sqrt 3 } \right)x = 15\sqrt 2 + 4\sqrt 3 \\
\Leftrightarrow x = \dfrac{{15\sqrt 2 + 4\sqrt 3 }}{{6\left( {\sqrt 2 + \sqrt 3 } \right)}} = \dfrac{{11\sqrt 6 – 18}}{6}\\
b)0 < x < 1 \Rightarrow \left\{ \begin{array}{l}
x > 0\\
1 – x > 0
\end{array} \right.\\
B = \dfrac{1}{{2x}} + \dfrac{1}{{1 – x}} = \dfrac{1}{{2x}} + \dfrac{2}{{2 – 2x}} = \dfrac{{{1^2}}}{{2x}} + \dfrac{{{{\sqrt 2 }^2}}}{{1 – 2x}} \ge \dfrac{{{{\left( {\sqrt 2 + 1} \right)}^2}}}{1} = \sqrt 2 + 1\\
\left( {B – C – S} \right)\\
\Rightarrow \min B = \sqrt 2 + 1\\
‘ = ‘ \Leftrightarrow \dfrac{1}{{2x}} = \dfrac{{\sqrt 2 }}{{2 – 2x}} \Leftrightarrow 2 – 2x = 2\sqrt 2 x\\
\Leftrightarrow x\left( {2 + 2\sqrt 2 } \right) = 2\\
\Leftrightarrow x = \sqrt 2 – 1
\end{array}$
$ => 2a^{2} = 3b^{2}$
Trên đây là hướng dẫn pp chọn điểm rơi