Cho hàm số y=(2x+4)/(x^2+4x+3) . Giải phương trình y''=0 .

Ta có:   y=2x+4x2+4x+3=2x+2x+22−1

⇒y'=2x+2x+22−1'=2x+22−2−2x+2.2x+2x+22−12=−2x+22−2x+22−12

⇒y''=−2x+22−2x+22−12'

=−4x+2x+22−12−−2x+22−2.2x+22−12x+2x+22−14

=4x+2x+22−1−x+22+1+2x+22+2x+22−14

=4x+2x+22−1x+22+3x+22−14

y''=0⇔4x+2x+22−1x+22+3x+22−14=0

x+22−1≠0

y''=0⇔x+2=0⇔x=−2