A = \(\left(m^2-4mp+4p^2\right)+10\left(m-2p\right)+25+\left(p^2-2p+1\right)+2\)

  \(=\left(m-2p\right)^2+2.5.\left(m-2p\right)+5^2+\left(p-1\right)^2+2\)

  \(=\left(m-2p+5\right)^2+\left(p-1\right)^2+2\ge2\)

Vậy: A min = 2 \(\Leftrightarrow m=-3;p=1\)

Ta có:\(p=\left(m^2-4mn+4n^2\right)+\left(10m-20n\right)+25+\left(n^2-2n+1\right)+6\)

\(\Rightarrow p=\left(m-2n\right)^2+2.5\left(m-2n\right)+5^2+\left(n-1\right)^2+6\)

\(\Rightarrow p=\left(m-2n+5\right)^2+\left(n-1\right)^2+6\ge6\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}m-2n+5=0\\n-1=0\end{cases}}\Rightarrow\hept{\begin{cases}m=-3\\n=1\end{cases}}\)

Vậy GTNN của p=6 khi m=-3  ;  n=1

Ta có: 

\(x^2-2\left(2m+1\right)x+4m^2+4m=0\\ \Leftrightarrow\left(x^2-2mx\right)-2\left(m+1\right)x+4m\left(m+1\right)=0\\ \Leftrightarrow x\left(x-2m\right)-2\left(m+1\right)\left(x-2m\right)=0\\ \Leftrightarrow\left(x-2m\right)\left(x-2m-2\right)=0\Leftrightarrow x_1=2m;...or...x_2=2m\)

\(\Rightarrow\left(x_1-2m\right)\left(x_2-2m\right)=0\Leftrightarrow\left(x_1-2m\right)^2\left(x_2-2m\right)^2=0\Leftrightarrow\left(x_1^2-4mx_1+4m^2\right)\left(x_2^2-4mx_2+4m^2\right)=0\)

\(A=4m^2+10m+9\)

\(A=4m^2+10m+\frac{25}{4}+\frac{11}{4}\)

\(A=\left(2m+\frac{5}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

Dấu "=" khi: \(m=-\frac{5}{4}\)

\(\left(2m+\frac{5}{4}\right)^2+\frac{11}{4}>\frac{11}{4}\)  thôi làm sao mà \(=\)  được hả bạn 

\(MinA=0\Leftrightarrow7+4x-4x^2=0\Leftrightarrow x=\dfrac{1\pm2\sqrt{2}}{2}\)

1) \(A=\frac{2018x^2-2.2018x+2018^2}{2018x^2}=\frac{\left(x-2018\right)^2+2017x^2}{2018x^2}=\frac{\left(x-2018\right)^2}{2018x^2}+\frac{2017}{2018}\)

vì \(\frac{\left(x-2018\right)^2}{2018x^2}\ge0\Rightarrow\frac{\left(x-2018\right)^2}{2018x^2}+\frac{2017}{2018}\ge\frac{2017}{2018}\)

dấu = xảy ra khi x-2018=0

=> x=2018

Vậy Min A=\(\frac{2017}{2017}\)khi x=2018

2) \(B=\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{3x^2+9x+7+10}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}=1+\frac{10}{3.x^2+9x+7}\)

\(=1+\frac{10}{3.\left(x^2+9x\right)+7}=1+\frac{10}{3.\left[x^2+\frac{2.x.3}{2}+\left(\frac{3}{2}\right)^2\right]-\frac{9}{4}+7}=1+\frac{10}{3.\left(x+\frac{9}{2}\right)^2+\frac{1}{4}}\)

để B lớn nhất => \(3.\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\)nhỏ nhất

mà \(3.\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)vì \(3.\left(x+\frac{3}{2}\right)^2\ge0\)

dấu = xảy ra khi \(x+\frac{3}{2}=0\)

=> x=\(-\frac{3}{2}\)

Vậy maxB=\(41\)khi x=\(-\frac{3}{2}\)

3) \(M=\frac{3x^2+14}{x^2+4}=\frac{3.\left(x^2+4\right)+2}{x^2+4}=3+\frac{2}{x^2+4}\)

để M lớn nhất => x²+4 nhỏ nhất

mà \(x^2+4\ge4\)(vì x² lớn hơn hoặc bằng 0)

dấu = xảy ra khi x² =0

=> x=0

Vậy Max M\(=\frac{7}{2}\)khi x=0

ps: bài này khá dài, sai sót bỏ qua =))

ê viết lộn dòng này :v

\(MinA=\frac{2017}{2018}\)nha 

Thanks. <3