Tìm Min,max: A=x^4+1/(x^2+1)^2 Question Tìm Min,max: A=x^4+1/(x^2+1)^2 in progress 0 Môn Toán Latifah 4 years 2020-11-01T03:31:37+00:00 2020-11-01T03:31:37+00:00 1 Answers 127 views 0
Answers ( )
*Min:
`A=(x^4+1)/((x^2+1)^2)=(2x^4+2)/(2(x^2+1)^2)=(x^4−2x^2+1+x^4+2x^2+1)/(2(x^2+1)2)`
`=((x^2−1)^2+(x^2+1)^2)/(2(x^2+1)^2)=((x^2−1)^2)/(2(x^2+1)^2)+1/2≥1/2`
Dấu ‘=’ xảy ra `⇔x=±1`
Vậy `Min=1/2` khi `x=+-1`
*Max:
`A=(x^4+1)/((x^2+1)^2)=(x^4+2x^2+1−2x^2)/((x^2+1)^2)=((x^2+1)^2−2x^2)/((x^2+1)^2)`
`=1−(2x^2)/((x^2+1)^2)≤1`
Dấu ‘=’ xảy ra `⇔x=0`
Vậy `Max=1 khi x=0`