Tìm Min P=\(\frac{2017-2015x}{\sqrt{1-x^2}}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\sqrt{1-x}=a\Rightarrow a^2=1-x\)
\(\sqrt{1+x}=b\Rightarrow b^2=1+x\)
\(\Rightarrow a^2+b^2=2\)
ta có: \(P=\frac{2017-2015x}{\sqrt{1-x^2}}=\frac{2015\left(1-x\right)+2}{\sqrt{\left(1-x\right)\left(1+x\right)}}\)
\(\Rightarrow P=\frac{2015.a^2+a^2+b^2}{ab}=\frac{2016a^2+b^2}{ab}\ge\frac{2.ab.\sqrt{2016}}{ab}=2\sqrt{2016}\)
=> GTNN của P là \(2\sqrt{2016}\)<=>\(a\sqrt{2016}=b\Leftrightarrow\sqrt{\left(1-x\right).2016}=\sqrt{1+x}\)
\(\Leftrightarrow x=\frac{2015}{2017}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{1}{2015}-\frac{2}{2015x}+\frac{1}{x^2}=\left(\frac{1}{x^2}-2.\frac{1}{x}.\frac{1}{2015}+\frac{1}{2015^2}\right)+\frac{1}{2015}-\frac{1}{2015^2}\)
\(=\left(\frac{1}{x}-\frac{1}{2015}\right)^2+\frac{2014}{2015^2}\ge\frac{2014}{2015^2}\)
\(MinP=\frac{2014}{2015^2}\) khi 1/x =1/2015 hay x = 2015
![](https://rs.olm.vn/images/avt/0.png?1311)
b)\(\frac{1}{a^2+a}=\frac{1}{a}.\frac{1}{a+1}=\frac{1}{a}\left(1-\frac{a}{a+1}\right)\ge\frac{1}{a}\left(1-\frac{\sqrt{a}}{2}\right)\)
\(=\frac{1}{a}-\frac{1}{2\sqrt{a}}\). Tương tự 2 BĐT còn lại và cộng theo vế thu được:
\(P\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)
\(\ge\frac{9}{a+b+c}-\frac{1}{2}.\frac{9}{\sqrt{a.1}+\sqrt{b.1}+\sqrt{c.1}}\)
\(\ge3-\frac{1}{2}.\frac{18}{a+b+c+3}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
Vậy..
\(x^2\ge0\)
\(\Rightarrow2015x\ge0\)
\(\Rightarrow1-x^2\ge1\)
\(\Rightarrow\sqrt{1-x^2}\ge1\)
\(\Rightarrow\dfrac{2017-2015x}{\sqrt{1-x^2}}\ge\dfrac{2017}{1}=2017\)
Dấu "=" xảy ra khi \(x^2=0\)
\(\Leftrightarrow x=0\)
Vậy \(P\min\limits=2017\Leftrightarrow x=0\)