Решить уравнение.а)×²-5√2×+12=0б)×²+2√3-72=0в)у²-6у+7=0г)р²-10р+7=0. Created by Dilyara555. algebra-ru

Решить уравнение.а)×²-5√2×+12=0б)×²+2√3-72=0в)у²-6у+7=0г)р²-10р+7=0...

August 2022 1 5 Report

Решить уравнение.
а)×²-5√2×+12=0
б)×²+2√3-72=0
в)у²-6у+7=0
г)р²-10р+7=0

Answers & Comments

alamode А)×²-5√2×+12=0
D=50-4*12=2
x1=(5√2-√2)/2
x2=(5√2+√2)/2

б)×²+2√3-72=0
D=12+4*72=300
x12=(-2√3+-√300)/2

в)у²-6у+7=0
D=36-4*7=4
y1=(6-2)/2=2
y2=(6+2)/2=4

г)р²-10р+7=0
D=100-4*7=68
x12=(10+-√68)/2

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Dilyara555 July 2022 | 0 Ответы

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