a) Ta có:
\(\dfrac{3x+7}{x^2+x+8}=\dfrac{x^2+x+8-x^2+2x-1}{x^2+x+8}\)
\(=1-\dfrac{x^2-2x+1}{x^2+x+8}=1-\dfrac{\left(x-1\right)^2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}}\)
Mà: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}>0\forall x\Rightarrow\dfrac{\left(x-1\right)^2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}}\ge0\forall x\)
\(\Rightarrow1-\dfrac{\left(x-1\right)^2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}}\le1\forall x\)
Dấu "=" xảy ra: \(\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy max của bt là 1 khi `x = 1`
b) Ta có:
\(\dfrac{2x-5}{x^2+4x+14}=\dfrac{-x^2-4x-14+x^2+6x+9}{x^2+4x+14}\)
\(=\dfrac{-\left(x^2+4x+14\right)}{x^2+4x+14}+\dfrac{x^2+6x+9}{x^2+4x+14}\)
\(=-1+\dfrac{\left(x+3\right)^2}{\left(x^2+2\cdot x\cdot2+2^2\right)+10}=-1+\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2+10}\)
Mà: \(\left(x+2\right)^2+10>0\forall x\Rightarrow\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2+10}\ge0\forall x\)
\(\Rightarrow-1+\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2+10}\ge-1\forall x\)
Dấu "=" xảy ra: \(\left(x+3\right)^2=0\Leftrightarrow x=-3\)
Vậy GTNN của bt là - 1 khi `x = -3`
