Question

3.Áp dụng bđt Cô-si, tìm GTNN:

a)\(y=\frac{x}{2}+\frac{2}{x-1};x>1\)

b)\(y=\frac{5x}{3}+\frac{5}{3x-1};x>\frac{1}{3}\)

c)\(y=\frac{2x}{1-x}+\frac{3}{x};0< x< 1\)

d)\(y=\frac{x^2+2020x+9}{x};x>0\)

Answer 1

\(y=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{2\left(x-1\right)}{2\left(x-1\right)}}+\frac{1}{2}=\frac{5}{2}\)

Dấu "=" xảy ra khi \(\frac{x-1}{2}=\frac{2}{x-1}\Rightarrow x=3\)

\(y=\frac{5\left(3x-1\right)}{9}+\frac{5}{3x-1}+\frac{5}{9}\ge2\sqrt{\frac{25\left(3x-1\right)}{9\left(3x-1\right)}}+\frac{5}{9}=\frac{35}{9}\)

Dấu "=" xảy ra khi \(x=\frac{4}{3}\)

\(y=-2+\frac{2}{1-x}+\frac{3}{x}\ge-2+\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{1-x+x}=3+2\sqrt{6}\)

Dấu "=" xảy ra khi \(\frac{1-x}{\sqrt{2}}=\frac{x}{\sqrt{3}}\Rightarrow x=3-\sqrt{6}\)

\(y=x+\frac{9}{x}+2020\ge2\sqrt{\frac{9x}{x}}+2020=2026\)

Dấu "=" xảy ra khi \(x=3\)