Câu 1:
\(\text{Ta có : }\dfrac{1}{B}=\dfrac{x-1}{x^2+8}\\ =\dfrac{8x-8}{8\left(x^2+8\right)}\\ =\dfrac{8x+8-16-x^2+x^2}{8\left(x^2+8\right)}\\ =\dfrac{-\left(x^2-8x+16\right)+\left(x^2+8\right)}{8\left(x^2+8\right)}\\ =\dfrac{-\left(x^2-8x+16\right)}{8\left(x^2+8\right)}+\dfrac{x^2+8}{8\left(x^2+8\right)}\\ =\dfrac{-\left(x-4\right)^2}{8\left(x^2+8\right)}+\dfrac{1}{8}\\ Do\text{ }-\left(x-4\right)^2\le0\forall x\\ \Rightarrow\dfrac{-\left(x-4\right)^2}{8\left(x^2+8\right)}\le0\forall x\\ \Rightarrow\dfrac{1}{B}=-\dfrac{\left(x-4\right)^2}{8\left(x^2+8\right)}+\dfrac{1}{8}\le\dfrac{1}{8}\forall x\\ \Rightarrow B\ge8\forall x\\ \text{Dấu "=" xảy ra khi: }\\ -\left(x-4\right)^2=0\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\\ \text{Vậy }B_{\left(Min\right)}=8\text{ khi }x=4\)
Câu 2:
\(A=\dfrac{2x^2+8x+15}{x^2+4x+7}\\ =\dfrac{2x^2+8x+14+1}{x^2+4x+7}\\ =\dfrac{\left(2x^2+8x+14\right)+1}{x^2+4x+7}\\ \\ =\dfrac{2\left(x^2+4x+7\right)}{x^2+4x+7}+\dfrac{1}{x^2+4x+7}\\ =2+\dfrac{1}{\left(x^2+4x+4\right)+3}\\ =2+\dfrac{1}{\left(x+2\right)^2+3}\\ \text{Do }\left(x+2\right)^2\ge0\forall x\\ \Rightarrow\left(x+2\right)^2+3\ge3\forall x\\ \Rightarrow\dfrac{1}{\left(x+2\right)^2+3}\ge\dfrac{1}{3}\forall x\\ \Rightarrow A=2+\dfrac{1}{\left(x+2\right)^2+3}\ge\dfrac{7}{3}\forall x\\ \text{ Dấu "=" xảy ra khi: }\\ \left(x+2\right)^2=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\\ \text{Vậy }A_{\left(Min\right)}=\dfrac{7}{3}\text{ }khi\text{ }x=-2\)
\(B=\dfrac{2x-1}{x^2}\\ \Rightarrow B-1=\dfrac{2x-1}{x^2}-1\\ \Rightarrow B-1=\dfrac{2x-1-x^2}{x^2}\\ \Rightarrow B-1=\dfrac{-\left(x^2-2x+1\right)}{x^2}\\ =\dfrac{-\left(x-1\right)^2}{x^2}\)
