Gọi \(A\) là tử (\(14x^2-8x+9\))

      \(C\) là mẫu (\(3x^2+6x+9\))

Ta có:\(A\) =\(14x^2-8x+9\)

\(\Rightarrow A_{min}\)=\(\frac{55}{7}\)

Ta có: \(C\)=\(3x^2+6x+9\)

\(\Rightarrow C_{min}\)=6

Suy ra \(B_{min}\)=\(\frac{\left(\frac{55}{7}\right)}{6}\)=\(\frac{55}{42}\)

Vậy GTNN của B là \(\frac{55}{42}\)

Ta có : \(B=\frac{14x^2-8x+9}{3x^2+6x+9}=\frac{2\left(x^2+2x+3\right)+\left(12x^2-12x+3\right)}{3\left(x^2+2x+3\right)}\)

\(=\frac{12\left(x-\frac{1}{2}\right)^2}{3\left(x^2+2x+3\right)}+\frac{2}{3}\ge\frac{2}{3}\) . Dấu "=" xảy ra khi x = 1/2

Vậy Min B = 2/3 khi x = 1/2

Tử=14(x-2/7)^2+55/7

Mẫu=3(x+1)^2+6

.... lm tiếp nhé mệt r

\(B=\frac{14\left(x^2+2x+3\right)-36x-33}{3\left(x^2+2x+3\right)}=\frac{14}{3}+\frac{-3.\left(12x+11\right)}{3.\left(x^2+2x+3\right)}=\frac{14}{3}-C\)

\(C=\frac{12x+11}{x^2+2x+3}=\frac{12\left(x+1\right)-1}{\left(x+1\right)^2+2}=\frac{12y-1}{y^2+2}=D\)

\(4-D=\frac{4y^2+8-\left(12y-1\right)}{4\left(y^2+2\right)}=\frac{\left(2y-3\right)^2}{4\left(y^2+2\right)}\ge0\)

\(D\le4\Rightarrow C\le4\Rightarrow B\ge\frac{14}{3}-4=\frac{2}{3}\)

GTNN B=2/3 khi y=3/2=> x=1/2

55/42

tớ thử rồi nhưng không phải

đơn giản