【在线等答案】解下列一元二次方程:3(2x-1)²=(2x-1)²+18;2x²-6x+7=0.
人气:248 ℃ 时间:2020-02-03 14:41:42
解答
3(2x-1)²=(2x-1)²
(2x-1)²(3-1)=2(2x-1)²=0
解得x=1/2
2x²-6x+7=0.
判别式=36-4×2×7=-20<0
所以此方程无实数根3(2x-1)²=(2x-1)²+18化为(2x-1)²(3-1)=2(2x-1)²=18(2x-1)²=92x-1=正负3x=2或x=-1
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