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Mình ko rõ đề bài
\(y=\frac{3x}{2}+\frac{1}{x}+1\)hay \(y=\frac{3x}{2}+\frac{1}{x+1}\)
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\(y=\frac{x}{3}+\frac{5}{2x-1}=\frac{2x}{6}+\frac{5}{2x-1}=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\)
\(\Rightarrow y\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=\frac{\sqrt{30}}{3}+\frac{1}{6}\)
\(\Rightarrow P_{min}=\frac{\sqrt{30}}{3}+\frac{1}{6}\)
Dấu "=" xảy ra khi \(\left(2x-1\right)^2=30\Rightarrow x=\frac{\sqrt{30}+1}{2}\)
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\(y=\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\)
\(\Rightarrow y\ge2\sqrt{\frac{3\left(x+1\right)}{2}.\frac{1}{x+1}}-\frac{3}{2}=\sqrt{6}-\frac{3}{2}\)
Dấu "=" khi \(\left(x+1\right)^2=\frac{2}{3}\Rightarrow x=\frac{\sqrt{6}}{3}-1\)
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Do \(x+3\) và \(5-2x\) đều không âm, áp dụng BĐT \(ab\le\frac{\left(a+b\right)^2}{4}\) ta có
\(y=\frac{1}{2}.\left(2x+6\right)\left(5-2x\right)\le\frac{1}{2}.\frac{\left(2x+6+5-2x\right)^2}{4}=\frac{1}{2}.\frac{11^2}{4}=\frac{121}{8}\)
\(\Rightarrow y_{max}=\frac{121}{8}\) khi \(2x+6=5-2x\Leftrightarrow x=\frac{-1}{4}\)
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Ta có : \(\dfrac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng )
Dấu "=" xảy ra khi \(a=b\)
Bài tập :
Có : \(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{x}+\dfrac{x+y}{y}=2+\dfrac{x}{y}+\dfrac{y}{x}\) ( do \(x+y=1\) )
Theo BĐT trên có : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2.\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)
Nên \(A\ge2+2=4\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
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Ta có : \(\dfrac{a+b}{2}\ge\sqrt{ab}\) (tự cm)
Lại có : \(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}\)
Áp dụng BĐT trên ta có : : \(xy\le\left(\dfrac{x+y}{2}\right)^2\)
\(\Leftrightarrow A\ge\dfrac{x+y}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{1}{\dfrac{1}{2^2}}=4\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy...
\(y=\frac{x}{1-x}+\frac{5}{x}=\frac{x}{1-x}+\frac{5\left(1-x\right)}{x}+5\)
Áp dụng BĐT Cô - si ta có :
\(\frac{x}{1-x}+\frac{5\left(1-x\right)}{x}\ge2\sqrt{\frac{5x\left(1-x\right)}{x\left(1-x\right)}}=2\sqrt{5}\)
\(\Rightarrow y\ge5+2\sqrt{5}\)
Dấu \("="\) xảy ra khi \(x^2=5x^2-10x+5\Leftrightarrow4x^2-10x+5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\frac{5+\sqrt{5}}{4}\\x_2=\frac{5-\sqrt{5}}{4}\end{matrix}\right.\)