\(S=\dfrac{2018x^2-2.2018x+2018^2}{2018x^2}=\dfrac{2017x^2+x^2-2.2018x+2018^2}{2018x^2}=\dfrac{2017}{2018}+\dfrac{\left(x-2018\right)^2}{x^2}\ge\dfrac{2017}{2018}\)

\(S_{min}=\dfrac{2017}{2018}\) khi \(x=2018\)

\(x+\dfrac{1}{x}=3\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27\\ \Leftrightarrow x^3+\left(\dfrac{1}{x}\right)^3+3x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}+3\cdot3=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}=18\)

b/ Ko biết yêu cầu

4/ \(E=\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{1}{x^3}+\frac{1}{x^3}\ge5\sqrt[5]{\frac{x^6}{27x^6}}=\frac{5}{\sqrt[5]{27}}\)

Dấu "=" xảy ra khi \(\frac{x^2}{3}=\frac{1}{x^3}\Leftrightarrow x=\sqrt[5]{3}\)

\(F=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge3\sqrt[3]{\frac{x^2}{4x^2}}=\frac{3}{\sqrt[3]{4}}\)

Dấu "=" xảy ra khi \(\frac{x}{2}=\frac{1}{x^2}\Rightarrow x=\sqrt[3]{2}\)

6/ \(Q=\frac{\left(x+1\right)^2+16}{2\left(x+1\right)}=\frac{x+1}{2}+\frac{8}{x+1}\ge2\sqrt{\frac{8\left(x+1\right)}{2\left(x+1\right)}}=4\)

Dấu "=" xảy ra khi \(\frac{x+1}{2}=\frac{8}{x+1}\Leftrightarrow x=3\)

7/

\(R=\frac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\ge2\sqrt{\frac{25\left(\sqrt{x}+3\right)}{\sqrt{x}+3}}=10\)

Dấu "=" xảy ra khi \(\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow x=4\)

8/

\(S=x^2+\frac{2000}{x}=x^2+\frac{1000}{x}+\frac{1000}{x}\ge3\sqrt[3]{\frac{1000^2x^2}{x^2}}=300\)

Dấu "=" xảy ra khi \(x^2=\frac{1000}{x}\Leftrightarrow x=10\)

\(\Delta=\left(-2m\right)^2-4\left(m^2-m+1\right)\)

=4m^2-4m^2+4m-4=4m-4

Để (1) có 2 nghiệm thì 4m-4>=0

=>m>=1