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\(P\left(x\right)=\frac{2012x+2013\sqrt{1-x^2}+2014}{\sqrt{1-x^2}}=\frac{2012x+2014}{\sqrt{1-x^2}}+\frac{2013\sqrt{1-x^2}}{\sqrt{1-x^2}}\)
\(=\frac{2012x+2014}{\sqrt{1-x^2}}+2013=2012+\frac{2012\left(1+x\right)+1-x}{\sqrt{1-x^2}}\)
Áp dụng BĐT AM-GM ta có:
\(P\left(x\right)\ge2012+\frac{2\sqrt{2012\left(1+x\right)\left(1-x\right)}}{\sqrt{1-x^2}}=2012+2\sqrt{2012}\)
=\(2013\) \(+\frac{2014+2012x}{\sqrt{1-x^2}}\) =\(\frac{2013\left(1+x\right)+1-x}{\sqrt{1-x^2}}\) \(\ge2013+\frac{2\sqrt{2013\left(1+x\right)\left(1-x\right)}}{\sqrt{1-x^2}}=2013+2\sqrt{2013}\)
dau = xay ra khi \(2013\left(1+x\right)=1-x\)
\(\Leftrightarrow x=-\frac{1001}{1002}\)
min p(x) =\(2013+2\sqrt{2013}\Leftrightarrow x=-\frac{1001}{1002}\)
a, Ta có : \(x=25\Rightarrow\sqrt{x}=\sqrt{25}=5\)
\(\Rightarrow Q=\frac{5-1}{5+1}=\frac{4}{6}=\frac{2}{3}\)
b, \(P=\frac{x\sqrt{x}-1}{x-\sqrt{x}}+\frac{x\sqrt{x}+1}{x+\sqrt{x}}-\frac{4}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{4}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1+x-\sqrt{x}+1-4}{\sqrt{x}}=\frac{2x-2}{\sqrt{x}}\)
c, Ta có : \(P.Q.\sqrt{x}< 8\)hay \(\frac{2x-2}{\sqrt{x}}.\sqrt{x}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)< 8\)
\(\Leftrightarrow\frac{2\left(x-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< 8\Leftrightarrow2\left(\sqrt{x}-1\right)^2< 8\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2< 4\Leftrightarrow\sqrt{x}-1< 2\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)
bài 1
a) ĐKXĐ : bạn tự tìm nhé
b) ta có A=\(\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)
=\(\sqrt{\left(\sqrt{x^2-1}+1\right)^2}+\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)
=\(\left|\sqrt{x^2-1}+1\right|+\left|\sqrt{x^2-1}-1\right|\)
=\(\sqrt{x^2-1}+1+\sqrt{x^2-1}-1\)( vì \(\left|x\right|\ge\sqrt{2}\))
=\(2\sqrt{x^2-1}\)
2. \(P=x^2-x\sqrt{3}+1=\left(x^2-x\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=\left(x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
Dấu '=' xảy ra khi \(x=\frac{\sqrt{3}}{2}\)
Vây \(P_{min}=\frac{1}{4}\)khi \(x=\frac{\sqrt{3}}{2}\)
3. \(Y=\frac{x}{\left(x+2011\right)^2}\le\frac{x}{4x.2011}=\frac{1}{8044}\)
Dấu '=' xảy ra khi \(x=2011\)
Vây \(Y_{max}=\frac{1}{8044}\)khi \(x=2011\)
4. \(Q=\frac{1}{x-\sqrt{x}+2}=\frac{1}{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{7}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}\le\frac{4}{7}\)
Dấu '=' xảy ra khi \(x=\frac{1}{4}\)
Vậy \(Q_{max}=\frac{4}{7}\)khi \(x=\frac{1}{4}\)