em chịu ạ! Tịt rùi! 

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

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

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

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

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

55/42

tớ thử rồi nhưng không phải

đơn giản

Xét biểu thức \(\frac{14x^2-8x+9}{3x^2+6x+9}=\frac{14x^2-8x+9}{3x^2+6x+9}-\frac{2}{3}+\frac{2}{3}=\frac{3\left(14x^2-8x+9\right)-2\left(3x^2+6x+9\right)}{3\left(3x^2+6x+9\right)}+\frac{2}{3}=\frac{36x^2-36x+9}{3\left(3x^2+6x+9\right)}+\frac{2}{3}=\frac{\left(6x-3\right)^2}{3\left(3x^2+6x+9\right)}+\frac{2}{3}\ge\frac{2}{3}\)Đẳng thức xảy ra khi x = ¹/₂

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

1, Ta có: \(A=3x^2+8x+9=3\left(x^2+\frac{8}{3}x+3\right)=3\left(x^2+\frac{8}{3}x+\frac{16}{9}+\frac{11}{9}\right)\)

\(=3\left(x+\frac{4}{3}\right)^2+\frac{11}{3}\ge\frac{11}{3}\forall x\)

=> Min A = 11/3 tại x = -4/3

2, Ta có: \(A=-2x^2+6x+3=-2\left(x^2-3x-\frac{3}{2}\right)=-2\left(x^2-3x+\frac{9}{4}-\frac{15}{4}\right)\)

\(=-2\left(x-\frac{3}{2}\right)^2+\frac{15}{2}\le\frac{15}{2}\forall x\)

=> Max A = 15/2 tại x = 3/2

=.= hk tốt!!

Cảm ơn

Tử=14(x-2/7)^2+55/7

Mẫu=3(x+1)^2+6

.... lm tiếp nhé mệt r