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Đặt \(t=\frac{x}{y}+\frac{y}{x};t\ne0\). Ta có:
\(t^2=\left(\frac{x}{y}+\frac{y}{x}\right)^2=\frac{x^2}{y^2}+2+\frac{y^2}{x^2}\)
\(\Rightarrow\frac{x^2}{y^2}+\frac{y^2}{x^2}=t^2-2\)
\(\Rightarrow P=t^2-2-t=\left(t-\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
Vậy GTNN của P là:\(-\frac{9}{4}\)khi \(t=\frac{1}{2}\)
P/s Các bạn tham khảo nha
1. x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)
2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)
3: \(A=\frac{x^2+x+4}{x+1}=\frac{\left(x^2+2x+1\right)-\left(x+1\right)+4}{x+1}=x+1-1+\frac{4}{x+1}\)
áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)
\(A-\frac{2013}{2014}=\frac{x^2-2x+2014}{x^2}-\frac{2013}{2014}=\frac{2014x^2-2.2014.x+2014^2-2013x^2}{2014x^2}\)
\(=\frac{x^2-2.x.2014+2014^2}{2014x^2}=\frac{\left(x-2014\right)^2}{2014x^2}\ge0\)
=>\(A\ge\frac{2013}{2014}\)
Dấu "=" xảy ra khi x=2014
Vậy minA=2013/2014 khi x=2014
A=\(\frac{2014x^2-2.2014x-2014^2}{2014x^2}\)=\(\frac{2013x^2+\left(x^2-2.2014x-2014^2\right)}{2014x^2}\)=\(\frac{2013x^2+\left(x-2014\right)^2}{2014x^2}\)=\(\frac{2013}{2014}+\frac{\left(x-2014\right)^2}{2014x^2}\ge\frac{2013}{2014}\)
vậy minA=\(\frac{2013}{2014}\)dấu bằng xảy ra khi x=2014
\(A=1-\frac{3}{x}+\frac{2019}{x^2}=2019\left(\frac{1}{x}-\frac{1}{1346}\right)^2+\frac{2689}{2692}\ge\frac{2689}{2692}\)
Dấu "=" xảy ra khi \(x=1346\)
\(A=1-\frac{2}{x}+\frac{2014}{x^2}\)
đặt 1/x=t ta có
\(A=1-2t+2014t^2\)
\(=2014\left(t^2-\frac{1}{1007}+\frac{1}{2014}\right)\)
=\(2014[\left(t-\frac{1}{2014}\right)^2-\left(\frac{1}{2014}\right)^2+\frac{1}{2014}]\)
=\(2014\left(t-\frac{1}{2014}\right)^2+\frac{2013}{2014}\)\(\ge\frac{2013}{2014}\)
dấu''='' xảy ra khi t-1/2014=0 <=>1/x=1/2014=>x=2014
\(B=\frac{x^2-2x+2018}{x^2}=\frac{2018x^2-2.2018.x+2018^2}{2018x^2}\)
\(=\frac{x^2-2.2018.x+2018^2}{2018x^2}+\frac{2017x^2}{2018x^2}\)
\(=\frac{\left(x-2018\right)^2}{x^2}+\frac{2017}{2018}\)
\(=\left(\frac{x-2018}{x}\right)^2+\frac{2017}{2018}\)
Vì : \(\left(\frac{x-2018}{x}\right)^2\ge0\forall x\)
Nên : \(B=\left(\frac{x-2018}{x}\right)^2+\frac{2017}{2018}\ge\frac{2017}{2018}\)
Vậy \(B_{min}=\frac{2017}{2018}\) khi x = 2018
\(\Leftrightarrow Bx^2-x^2+2x-2018=0\)
\(\Leftrightarrow\left(B-1\right)x^2+2x-2018=0\)
Để tồn tại x thì \(\Delta^'\ge0\)
\(\Leftrightarrow1+2018\left(B-1\right)\ge0\)
\(\Leftrightarrow B\ge\frac{2017}{2018}\)
Vậy MinB=2017/2018, dấu bằng xảy ra khi x=2018
\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)\ge\frac{1}{2}.2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}=x^4y^4\)
\(x^{16}+y^{16}+1+1+1+1+1+1\ge8\sqrt[8]{x^{16}y^{16}}=8x^2y^2\)
\(\Rightarrow A\ge x^4y^4+\frac{1}{4}\left(8x^2y^2-6\right)-\left(x^4y^4+2x^2y^2+1\right)=-\frac{5}{2}\)
Dấu "=" xảy ra khi \(x^2=y^2=1\)
Vậy GTNN của A là -5/2.
\(A=1-\frac{2}{x}+\frac{2014}{x^2}\)\(=\left(\frac{1}{x}-\frac{1}{2014}\right)^2+\frac{2013}{2014}\ge\frac{2013}{2014}\)
\(A_{min}=\frac{2013}{2014}\Leftrightarrow x=2014\left(TM\right)\)