y′=(2x+1)′(x–3)–(x–3)′(2x+1)(x–3)2=–7(x–3)2 20/08/2021 Bởi Arya y′=(2x+1)′(x–3)–(x–3)′(2x+1)(x–3)2=–7(x–3)2
Đáp án: y′=(2x+1)′(x−3)−(x−3)′(2x+1).(x−3)2y′=(2x+1)′(x−3)−(x−3)′(2x+1).(x−3)2 =2(x−3)−(2x+1)(x−3)=2(x−3)−(2x+1)(x−3) =(x−3)(2−2x−1)=(x−3)(2−2x−1) =(x−3)(−2x+1)=(x−3)(−2x+1) =−2x2+7x−3 WODVoHoa Bình luận
`y′=(2x+1)′(x−3)−(x−3)′(2x+1)(x−3)^2``=2(x−3)−(2x+1)(x−3)=2(x−3)−(2x+1)(x−3)``=(x−3)(2−2x−1)=(x−3)(2−2x−1)``=(x−3)(−2x+1)``=(x−3)(−2x+1)`$=\left\{\matrix{x-3=0\hfill\ \cr -2x+1=0\hfill}\right.$$=\left\{\matrix{x=3\hfill\ \cr -2x=-1\hfill}\right.$$=\left\{\matrix{x=3\hfill\ \cr x=1/2\hfill}\right.$ $#Blink$ $\boxed{\text{@Rosé}}$ Bình luận
Đáp án:
y′=(2x+1)′(x−3)−(x−3)′(2x+1).(x−3)2y′=(2x+1)′(x−3)−(x−3)′(2x+1).(x−3)2
=2(x−3)−(2x+1)(x−3)=2(x−3)−(2x+1)(x−3)
=(x−3)(2−2x−1)=(x−3)(2−2x−1)
=(x−3)(−2x+1)=(x−3)(−2x+1)
=−2x2+7x−3
WODVoHoa
`y′=(2x+1)′(x−3)−(x−3)′(2x+1)(x−3)^2`
`=2(x−3)−(2x+1)(x−3)=2(x−3)−(2x+1)(x−3)`
`=(x−3)(2−2x−1)=(x−3)(2−2x−1)`
`=(x−3)(−2x+1)`
`=(x−3)(−2x+1)`
$=\left\{\matrix{x-3=0\hfill\ \cr -2x+1=0\hfill}\right.$
$=\left\{\matrix{x=3\hfill\ \cr -2x=-1\hfill}\right.$
$=\left\{\matrix{x=3\hfill\ \cr x=1/2\hfill}\right.$
$#Blink$ $\boxed{\text{@Rosé}}$