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GameKyu99
@GameKyu99
July 2022
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CРОЧНО! ПОМОГИТЕ РЕШИТЬ!
Решите уравнение:
[tex]f'(x)=0, [/tex] если [tex]f'(x)=0, f(x) = \frac{0,5x^2- x+1}{0,5x-0,5} [/tex]
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sedinalana
Verified answer
F(x)=(0,5x²-x+1)/(0,5x-0,5)=0,5(x²-2x+2)/0,5(x-1)=(x²-2x+2)/(x-1)
f`(x)=[(2x-2)(x-1)-1(x²-2x+2)]/(x-1)²=(2x²-4x+2-x²+2x-2)/(x-1)²=
=(x²-2x)/(x-1)²=0
x(x-2)=0
x=0
x=2
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Answers & Comments
Verified answer
F(x)=(0,5x²-x+1)/(0,5x-0,5)=0,5(x²-2x+2)/0,5(x-1)=(x²-2x+2)/(x-1)f`(x)=[(2x-2)(x-1)-1(x²-2x+2)]/(x-1)²=(2x²-4x+2-x²+2x-2)/(x-1)²=
=(x²-2x)/(x-1)²=0
x(x-2)=0
x=0
x=2