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 

Bài 2:

\(A=x^2+2x+2012\)

 \(=\left(x^2+2x+1\right)+2011\)

\(=\left(x+1\right)^2+2011\)

Ta có: \(\left(x+1\right)^2\ge0,\forall x\)

\(\Rightarrow\left(x+1\right)^2+2011\ge2011,\forall x\)

Hay \(A\ge2011,\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy Min A=2011 tại x=-1

làm chuẩn đấy

a)

\(\dfrac{H}{x^2+9x+14}=\dfrac{1-x}{x+2}\)

\(\Rightarrow\dfrac{H}{x^2+7x+2x+14}=\dfrac{1-x}{x+2}\)

\(\Rightarrow\dfrac{H}{\left(x+7\right)\left(x+2\right)}=\dfrac{1-x}{x+2}\)

\(\Rightarrow\left(x+2\right)\left(x+7\right)\left(1-x\right)=H.\left(x+2\right)\)

\(\Rightarrow H=\left(x+7\right)\left(1-x\right)\)

b)

\(\dfrac{2x^2-5x+2}{x^2+5x-14}=\dfrac{2x-1}{H}\)

\(\Rightarrow\dfrac{2x^2-4x-x+2}{x^2+7x-2x-14}=\dfrac{2x-1}{H}\)

\(\Rightarrow\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x+7\right)\left(x-2\right)}=\dfrac{2x-1}{H}\)

\(\Rightarrow\left(2x-1\right)\left(x-2\right).H=\left(2x-1\right)\left(x+7\right)\left(x-2\right)\)

\(\Rightarrow H=x+7\)

a: \(=\dfrac{4}{x+2}-\dfrac{3}{x-2}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

b: \(=\dfrac{6x+3\left(x-1\right)+2\left(x-2\right)}{6}=\dfrac{6x+3x-3+2x-4}{6}=\dfrac{11x-7}{6}\)

c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)

Ta có

\(\frac{a}{x}+\frac{a+1}{x-1}+\frac{a+2}{x-2}=\frac{\left(3a+3\right)x^2-\left(6a+4\right)x+2a}{x^3-3x^2+2x}=\frac{9x^2-16x+4}{x^3-3x^2+2x}\)

Dấu = xảy ra khi

\(\hept{\begin{cases}3a+3=9\\-6a-4=-16\\2a=4\end{cases}\Leftrightarrow a=2}\)