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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

Tử=14(x-2/7)^2+55/7

Mẫu=3(x+1)^2+6

.... lm tiếp nhé mệt r

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

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

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