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数学
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x²+(√5+√3)x+√15=0 (x²+x)(x²+x—2)=24 (x—√2)=√5x(2—√2x)
解一元二次方程
人气:399 ℃ 时间:2020-04-20 02:36:52
解答
1题x²+(√5+√3)x+√15=0 x²+(√5+√3)x+√5*√3=0(x+√5)(x+√3)=0x1=-√5x2=-√32题(x²+x)(x²+x-2)=24 (x²+x)²-2(x²+x)-24=0(x²+x+4)(x²+x-6)=0[(x+1/2)²+3/4)...
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