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VikKod
@VikKod
October 2021
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Разложите на множители квадратный трехчлен:
а) x^2-10x+21
б)5y^2+9y-2
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tamarabernukho
x^2-10x+21
D=100-4*21=100-84=16
x1=(10+4)/2=7
x2=(10-4)/2=3
x
²-10x+21=(x-7)(x-3)
5y^2+9y-2
D=81+4*5*2=81+40=121
y1=(-9-11)/10=-2
y2=(-9+11)10=1/5
5y²+9y-2=(y+2)(y-1/5)*5=(y+2)(5y-1)
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Answers & Comments
D=100-4*21=100-84=16
x1=(10+4)/2=7
x2=(10-4)/2=3
x²-10x+21=(x-7)(x-3)
5y^2+9y-2
D=81+4*5*2=81+40=121
y1=(-9-11)/10=-2
y2=(-9+11)10=1/5
5y²+9y-2=(y+2)(y-1/5)*5=(y+2)(5y-1)