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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

55/42

tớ thử rồi nhưng không phải

đơn giản

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

\(B=\dfrac{2}{3}+\dfrac{\left(2x-1\right)^2}{x^2+2x+3}=\dfrac{2}{3}+\dfrac{\left(2x-1\right)^2}{\left(x+1\right)^2+2}\)

vì \(\dfrac{\left(2x-1\right)^2}{\left(x+1\right)^2+2}\ge0\)

\(B_{nn}\Leftrightarrow\dfrac{\left(2x-1\right)^2}{\left(x+1\right)^2+2}\)(nn)

\(\Rightarrow B\ge\dfrac{2}{3}\)

B(nn)=\(\dfrac{2}{3}\) ; khi 2x-1 =0 hay x=1/2

Min_B=2/3 khi x=1/2

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1