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\(y=\frac{2x+1}{x^2+2}\)
\(\Rightarrow y+\frac{1}{2}=\frac{2x+1}{x^2+2}+\frac{1}{2}\)
\(=\frac{2\left(2x+1\right)+x^2+2}{2\left(x^2+2\right)}\)
\(=\frac{4x+2+x^2+2}{2\left(x^2+2\right)}\)
\(=\frac{\left(x+2\right)^2}{2\left(x^2+2\right)}\)
Vì \(\left(x+2\right)^2\ge0\) với mọi x
\(2\left(x^2+2\right)\ge0\) với mọi x
\(\Rightarrow y+\frac{1}{2}\ge0\)
\(\Rightarrow y\ge-\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy GTNN của \(y=-\frac{1}{2}\) tại \(x=-2\)
\(y=\dfrac{x+3}{4}+\dfrac{9}{x-1}=\dfrac{x-1}{4}+\dfrac{9}{x-1}+1\)
\(y\ge2\sqrt{\dfrac{9\left(x-1\right)}{4\left(x-1\right)}}+1=4\)
\(y_{min}=4\) khi \(x=7\)
Ta có :
\(y=\frac{2}{1-x}+\frac{1}{x}\)
\(\Rightarrow y=\frac{2\left(1-x\right)+2x}{1-x}+\frac{1-x+x}{x}\)
\(\Rightarrow y=2+\frac{2x}{1-x}+\frac{1-x}{x}+1\)
\(\Rightarrow y=\frac{2x}{1-x}+\frac{1-x}{x}+3\)
Vì \(0< x< 1\Rightarrow\hept{\begin{cases}\frac{2x}{1-x}>0\\\frac{1}{x}>0\end{cases}}\)
Áp dụng BĐT Cô si cho 2 số dương , ta có :
\(\Rightarrow y=\frac{2x}{1-x}+\frac{1-x}{x}+3\ge2\sqrt{\frac{2x}{1-x}.\frac{1-x}{x}}+3=2\sqrt{2}+3\)
Dấu "=" xảy ra khi \(\frac{2x}{1-x}=\frac{1-x}{x}\Leftrightarrow\left(1-x\right)^2=2x^2\Leftrightarrow x^2+2x-1=0\Leftrightarrow\left(x+1\right)^2=2\Rightarrow x=\sqrt{2}-1\)
( vì\(0< x< 1\) )
Vậy \(Min_y=2\sqrt{2}+3\) khi \(x=\sqrt{2}-1\)
\(y=\frac{2}{1-x}+\frac{1}{x}\ge\frac{\left(\sqrt{2}+1\right)^2}{1-x+x}=3+2\sqrt{2}\)
Dấu = xảy ra khi
\(\frac{\sqrt{2}}{1-x}=\frac{1}{x}\)
\(\Leftrightarrow x=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
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: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)
\(Y=\frac{x^2+x+1}{x^2+2x+2}=1-\frac{x+1}{x^2+2x+2}.Y_{min}\Leftrightarrow\frac{x+1}{x^2+2x+2}.Dat:GTLN\)
\(1-\frac{x+1}{x^2+2x+2}\ge\frac{1}{2}\)
Dấu "=" xảy ra khi:
x=0