|3x-7|+|3x-2|+8 >= 5+8 = 13 

Dấu "=" xảy ra <=> 3/2 <= x <= 7/3

k mk nha

tiếp đi bạn 

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

\(\Leftrightarrow Px^2+2P=3x^2-4x+8\)

\(\Leftrightarrow\left(P-3\right)x^2+4x+2P-8=0\)

\(\Delta'=4-\left(P-3\right)\left(2P-8\right)\ge0\)

\(\Rightarrow-P^2+7P-10\ge0\)

\(\Rightarrow2\le P\le5\)

\(P_{min}=2\) khi \(x=2\)

\(P_{max}=5\) khi \(x=-1\)

1) Tìm GTNN:

\(P=\frac{3x^2-4x+8}{x^2+2}=\frac{2\left(x^2+2\right)+x^2-4x+4}{x^2+2}=2+\frac{\left(x-2\right)^2}{x^2+2}\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(P_{Min}=2\) khi \(x=2\)

2) Tìm GTLN:

\(P=\frac{3x^2-4x+8}{x^2+2}=\frac{5\left(x^2+2\right)-2x^2-4x-2}{x^2+2}=5-\frac{2\left(x+1\right)^2}{x^2+2}\le5\)

Dấu "=" xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy \(P_{Max}=5\) khi \(x=-1\)

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

\(=3x^2-5x+\frac{3}{2}\)

\(=3\left(x^2-2\cdot\frac{5}{6}\cdot x+\frac{25}{36}\right)+\frac{1}{4}\)

\(=3\left(x-\frac{5}{6}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

Đẳng thức xảy ra tại \(x=\frac{5}{6}\)