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\(y=\dfrac{x}{2}+\dfrac{18}{x}\ge2\sqrt{\dfrac{18x}{2x}}=6\)
\(y_{min}=6\) khi \(x=6\)
Theo đề: 0 < x < 1 => \(\left\{{}\begin{matrix}\frac{4}{x}>0\\\frac{9}{1-x}>0\end{matrix}\right.\)
⇔A = \(\frac{4}{x}\)+ \(\frac{9}{1-x}\) ≥ \(\frac{\left(2+3\right)^2}{x+1-x}\)= 25
Dấu "=" xảy ra ⇔ 9x = 4(1 - x) ⇔ x =\(\frac{2}{5}\) (TM)
a/ \(\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}.\frac{18}{x}}=...\)
b/ \(\frac{x}{2}+\frac{2}{x-1}=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{x-1}{2}.\frac{2}{x-1}}+\frac{1}{2}=...\)
c/ \(\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2}.\frac{1}{x+1}}-\frac{3}{2}=...\)
d/ \(\frac{x}{3}+\frac{5}{2x-1}=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=...\)
e/ \(\frac{x}{1-x}+\frac{5}{x}=\frac{x}{1-x}+\frac{5-5x+5x}{x}=\frac{x}{1-x}+\frac{5\left(1-x\right)}{x}+5\ge2\sqrt{\frac{x}{1-x}.\frac{5\left(1-x\right)}{x}}+5=...\)
f/ \(\frac{x^3+1}{x^2}=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge2\sqrt{\frac{x}{2}.\frac{x}{2}.\frac{1}{x^2}}=...\)
g/ \(\frac{x^2+4x+4}{x}=x+\frac{4}{x}+4\ge2\sqrt{x.\frac{4}{x}}+4=...\)
Ta có : \(4P=\frac{16}{x}+\frac{1}{y}\ge\frac{\left(4+1\right)^2}{x+y}=\frac{25}{\frac{5}{4}}=20\)
\(\Rightarrow P\ge5\)
Dấu \("="\) xảy ra khi : \(\left\{{}\begin{matrix}x+y=\frac{5}{4}\\\frac{4}{x}=\frac{1}{y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{1}{4}\end{matrix}\right.\)
Mình áp dụng luôn Cô - si cho các số ta được
a) \(\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}\cdot\frac{18}{x}}=2.\sqrt{9}=2.3=6\)
b) \(y=\frac{x}{2}+\frac{2}{x-1}=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{x-1}{2}\cdot\frac{2}{x-1}}+\frac{1}{2}=2+\frac{1}{2}=\frac{5}{2}\)
c) \(\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2}\cdot\frac{1}{x+1}}-\frac{3}{2}=2\sqrt{\frac{3}{2}}-\frac{3}{2}=\frac{-3+2\sqrt{6}}{2}\)
h) \(x^2+\frac{2}{x^2}\ge2\sqrt{x^2\cdot\frac{2}{x^2}}=2\sqrt{2}\)
g) \(\frac{x^2+4x+4}{x}=\frac{\left(x+2\right)^2}{x}\ge0\)
\(P=\frac{16}{x}+\frac{\frac{1}{4}}{y}\ge\frac{\left(4+\frac{1}{2}\right)^2}{x+y}=\frac{81}{20}\)
\(\Rightarrow a+b=81+20=101\)
(Mấy nay cứ thấy bài này nó cập nhật lại mãi :D)
Bđt Cauchy- Schwars (vì 0<x<1)
\(\frac{4}{x}+\frac{1}{1-x}\ge\frac{\left(2+1\right)^2}{x+1-x}=9\)
Dấu "=" xảy ra
<=> \(\frac{2}{x}=\frac{1}{1-x}\Leftrightarrow x=\frac{2}{3}\)
\(A=\frac{4}{x}+\frac{\frac{1}{4}}{y}\ge\frac{\left(2+\frac{1}{2}\right)^2}{x+y}=5\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=1\\y=\frac{1}{4}\end{matrix}\right.\)
Lam ro ra mot chut dc k ban minh k hieu gi ca