Áp dụng bất đẳng thức Svacxo, ta có:

\(f\left(x\right)=\dfrac{4}{x}+\dfrac{9}{1-x}=\dfrac{2^2}{x}+\dfrac{3^2}{1-x}\ge\dfrac{\left(2+3\right)^2}{x+1-x}=25\)

Vậy \(f\left(x\right)_{min}=25\Leftrightarrow\dfrac{2}{x}=\dfrac{3}{1-x}\Leftrightarrow x=\dfrac{2}{5}\)

\(P=\dfrac{4}{x}+1+\dfrac{9}{1-x}=\dfrac{4}{x}+25x+25\left(1-x\right)+\dfrac{9}{1-x}-24\)

\(\Rightarrow P\ge2\sqrt{\dfrac{4}{x}.25x}+2\sqrt{25\left(1-x\right).\dfrac{9}{1-x}}-24\)

\(\Rightarrow P\ge20+30-24=26\)

\(\Rightarrow P_{min}=26\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{4}{x}=25x\\25\left(1-x\right)=\dfrac{9}{1-x}\end{matrix}\right.\) \(\Rightarrow x=\dfrac{2}{5}\)

\(1+2x+\dfrac{4}{1+2x}-1\ge4-1=3\)

khi 2x+1=2hay x=1/2

a) \(\dfrac{\left(x-1\right)^2}{x-2}=\dfrac{\left(x-2\right)^2+2\left(x-2\right)+1}{x-2}=x-2+2+\dfrac{1}{x-2}\ge2+2\sqrt{\left(x-2\right).\dfrac{1}{x-2}}=4\)

GTNN là 4 khi x=3

Lời giải:

Ta có:

\(f(x)=\frac{x}{2}+\frac{1}{2x+1}=\frac{2x}{4}+\frac{1}{2x+1}=\frac{2x+1}{4}+\frac{1}{2x+1}-\frac{1}{4}\)

Vì \(x>\frac{-1}{2}\Rightarrow 2x+1>0\). Áp dụng BĐT Cauchy cho các số dương ta có:

\(\frac{2x+1}{4}+\frac{1}{2x+1}\geq 2\sqrt{\frac{2x+1}{4}.\frac{1}{2x+1}}=1\)

\(\Rightarrow f(x)=\frac{2x+1}{4}+\frac{1}{2x+1}-\frac{1}{4}\ge 1-\frac{1}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi \(\frac{2x+1}{4}=\frac{1}{2x+1}\Leftrightarrow x=\frac{1}{2}\)

Vậy GTNN của \(f(x)=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)