Bài 10:
\(B=\dfrac{x^2-1}{x^2+1}\)
=>\(B=\dfrac{x^2+1-2}{x^2+1}=1-\dfrac{2}{x^2+1}\)
\(x^2+1>=1\forall x\)
=>\(\dfrac{2}{x^2+1}< =\dfrac{2}{1}=2\forall x\)
=>\(-\dfrac{2}{x^2+1}>=-2\forall x\)
=>\(B=\dfrac{-2}{x^2+1}+1>=-2+1=-1\forall x\)
Dấu '=' xảy ra khi x=0
Vậy: \(B_{min}=-1\) khi x=0
Bài 9:
\(A=\dfrac{1}{\left|x+2017\right|+\left|x-2\right|}\)
\(\left|x+2017\right|+\left|x-2\right|=\left|x+2017\right|+\left|2-x\right|>=\left|x+2017+2-x\right|=2019\)
=>\(A=\dfrac{1}{\left|x+2017\right|+\left|x+2\right|}< =\dfrac{1}{2019}\forall x\)
Dấu '=' xảy ra khi \(\left(x+2017\right)\left(x-2\right)< =0\)
=>-2017<=x<=2
Đúng 2
Bình luận (0)
