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Iwillims
@Iwillims
July 2022
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Решите систему неравенств x^+4x <1
x^2+4x>-1
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miaxx
Х²+4х
<1
x
²+4x-1<0
x²+
4x-1=0
D=16-4*1*(-1)=16+4=20 √d= 2√5
x1= -4+2√5/2=-√5
x2=-4-2√5/2=-3√5
x²+4>-1
x²+4+1>0
x²+4+1=0
D=16-4*1*1=16-4=12√D=2√3
x1=-4+2√3/2=-√3
x2=-4-2√3/2=-3√3
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Answers & Comments
x²+4x-1<0
x²+4x-1=0
D=16-4*1*(-1)=16+4=20 √d= 2√5
x1= -4+2√5/2=-√5
x2=-4-2√5/2=-3√5
x²+4>-1
x²+4+1>0
x²+4+1=0
D=16-4*1*1=16-4=12√D=2√3
x1=-4+2√3/2=-√3
x2=-4-2√3/2=-3√3