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一元二次方程y²-5y-6=0 x(x+1)-5x=0 (x+2)(x-2)=1 8y²-2=4y (x-3)(x-2)=42 4(x-1)²-x+1=0
人气:211 ℃ 时间:2019-11-09 07:16:22
解答
.y²-5y-6=0
(y-6)(y+1)=0
y1=6 y2=-1
2.(x+2)(x-2)=1
x²-5=0
x1=√5 x2=-√5
3.8y²-2=4y
4y²-2y-1=0
Δ=4+16=20
y1=(1+√5)/4 y2=(1-√5)/4
4.(x-3)(x-2)=42
x²-5x-36=0
(x-9)(x+4)=0
x1=9 x2=-4
5.4(x-1)²-x+1=0
4x²-9x+5=0
(4x-5)(x-1)=0
x1=5/4 x2=1
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