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解下列一元二次方程:(1)9(x-2)²-121=0 (2)2(x-3)²=x(x-3)
(1)9(x-2)²-121=0
(2)2(x-3)²=x(x-3)
(3)x²-2倍的根号5x+1=0
(4)(x-1)²-5(x-1)+6=0
人气:122 ℃ 时间:2019-12-09 19:29:45
解答
(1)9(x-2)²=121
(x-2)²=121/9
x-2=±11/3
x=11/3+2或x=2-11/3
x=17/3或x=-5/3
(2)2x²-12x+18=x²-3x
x²-9x+18=0
(x-3)(x-6)=0
x=3或x=6
(3)(x²-2)²(5x+1)=0
x²-2=0或5x+1=0
x等于正负根号2或x=-0.2
(4)[(x-1)-2][(x-1)-3]=0
(x-3)(x-4)=0
x=3或x=4
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