bạn nói với mình điều kiện x>2 vậy làm như sau:

Đặt:\(A=\frac{3x-x^2-18}{x-2}=-\frac{x^2-3x+18}{x-2}=-\frac{x^2-4x+4+x-2+16}{x-2}\)

\(=-\frac{\left(x-2\right)^2+\left(x-2\right)+16}{x-2}\)\(=-\left(x-2+1+\frac{16}{x-2}\right)\)

Áp dụng bđt Cô si cho 2 số dương ta được: \(x-2+\frac{16}{x-2}\ge2\sqrt{\left(x-2\right).\frac{16}{x-2}}=8\)

=>\(x-2+\frac{16}{x-2}+1\ge9\)=>\(A=-\left(x-2+1+\frac{16}{x-2}\right)\le-9\)

=> maxA=-9 <=> x=6

A = \(\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{1}{x^3}+\frac{1}{x^3}\ge5\sqrt[5]{\frac{1}{27}}\)

dấu bằng xảy ra khi x = \(\sqrt[5]{3}\)

P=1+2x/(x^2+1)<= 1+(x^2+1)/(x^2+1)=2
Suy ra Pmax =2 khi x=1

\(x+\dfrac{16}{x-1}\\ =x-1+\dfrac{16}{x-1}+1\)

Áp dụng BĐT Cô-si ta có:
\(x-1+\dfrac{16}{x-1}+1\\ \ge2\sqrt{\left(x-1\right).\dfrac{16}{x-1}}+1\\ =2\sqrt{16}+1\\ =9\)

Dấu "=" xảy ra

 \(\Leftrightarrow x-1=\dfrac{16}{x-1}\\ \Leftrightarrow\left(x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Mình ko rõ đề bài 

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