Câu 5:
a: y'=2*5x^4-3*2x=10x^4-6x
f'(-2)=10*16-6*(-2)=160+12=172
f(-2)=2*(-32)-3*4=-64-12=-76
y-f(-2)=f'(-2)(x+2)
=>y+76=172(x+2)
=>y=172x+344-76=172x+268
b: \(y'=\dfrac{\left(3-2x\right)'\left(x+3\right)-\left(3-2x\right)\left(x+3\right)'}{\left(x+3\right)^2}=\dfrac{-2\left(x+3\right)-\left(3-2x\right)}{\left(x+3\right)^2}=\dfrac{-2x-6-3+2x}{\left(x+3\right)^2}\)
=-9/(x+3)^2
(d) vuông góc -9x+y-3=0
nên (d): x+9y+c=0
=>9y=-x-c
=>y=-1/9x-c/9
=>f'(x)=-1/9
=>(x+3)^2=81
=>x=6 hoặc x=-12
Khi x=6 thì f(6)=-1
Khi x=-12 thì f(-12)=-3
TH1: x=6; f(6)=-1; f'(6)=-1/9
y-f(6)=f'(6)(x-6)
=>y+1=-1/9(x-6)=-1/9x+2/3
=>y=-1/9x-1/3
TH2: x=-12; f(-12)=-3; f'(-12)=-1/9
y-f(-12)=f'(-12)(x+12)
=>y+3=-1/9(x+12)=-1/9x-4/3
=>y=-1/9x-13/3