Mình ko rõ đề bài 

\(y=\frac{3x}{2}+\frac{1}{x}+1\)hay \(y=\frac{3x}{2}+\frac{1}{x+1}\)

\(y=\frac{x}{3}+\frac{5}{2x-1}=\frac{2x}{6}+\frac{5}{2x-1}=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\)

\(\Rightarrow y\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=\frac{\sqrt{30}}{3}+\frac{1}{6}\)

\(\Rightarrow P_{min}=\frac{\sqrt{30}}{3}+\frac{1}{6}\)

Dấu "=" xảy ra khi \(\left(2x-1\right)^2=30\Rightarrow x=\frac{\sqrt{30}+1}{2}\)

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

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 \(x^2=5x^2-10x+5\Leftrightarrow4x^2-10x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\frac{5+\sqrt{5}}{4}\\x_2=\frac{5-\sqrt{5}}{4}\end{matrix}\right.\)

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

\(\Rightarrow y\ge2\sqrt{\frac{3\left(x+1\right)}{2}.\frac{1}{x+1}}-\frac{3}{2}=\sqrt{6}-\frac{3}{2}\)

Dấu "=" khi \(\left(x+1\right)^2=\frac{2}{3}\Rightarrow x=\frac{\sqrt{6}}{3}-1\)

Đề sai, cho đk x mà ko có đk y sao áp dụng cauchy bây giờ:v

\(\dfrac{x+5}{\sqrt{x}+2}\) + 11= \(\dfrac{x-4}{\sqrt{x}+2}\)+\(\dfrac{9}{\sqrt{x}+2}\)+11=\(\sqrt{x}\)-2+11+\(\dfrac{9}{\sqrt{x}+2}\)=\(\sqrt{x}\)+2+\(\dfrac{9}{\sqrt{x}+2}\)+9

lớn hơn hoặc bằng 15 khi và chỉ khi x=3

       Câu b bn giải tương tự nhé