北师大版初一下数学,不能用初三的知识!
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/27 06:26:10
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![北师大版初一下数学,不能用初三的知识!](/uploads/image/z/16587187-43-7.jpg?t=%E5%8C%97%E5%B8%88%E5%A4%A7%E7%89%88%E5%88%9D%E4%B8%80%E4%B8%8B%E6%95%B0%E5%AD%A6%2C%E4%B8%8D%E8%83%BD%E7%94%A8%E5%88%9D%E4%B8%89%E7%9A%84%E7%9F%A5%E8%AF%86%21)
x²+3x-2=0.(1);y²+3y-2=0.(2)
(1)-(2)得x²-y²+3(x-y)=0
分解因式得 (x+y)(x-y)+3(x-y)=0
提公因式得 (x-y)[(x+y)+3]=0
因为x≠y,即x-y≠0,故必有x+y+3=0,即有x+y=-3.(3)
(1)+(2)得x²+y²+3(x+y)-4=x²+y²-9-4=x²+y²-13=0,故x²+y²=13.(4)
将(3)两边平方得x²+y²+2xy=9,故xy=[9-(x²+y²)]/2=(9-13)/2=-2
∴(y/x)+(x/y)=(x²+y²)/xy=-13/2
(1)-(2)得x²-y²+3(x-y)=0
分解因式得 (x+y)(x-y)+3(x-y)=0
提公因式得 (x-y)[(x+y)+3]=0
因为x≠y,即x-y≠0,故必有x+y+3=0,即有x+y=-3.(3)
(1)+(2)得x²+y²+3(x+y)-4=x²+y²-9-4=x²+y²-13=0,故x²+y²=13.(4)
将(3)两边平方得x²+y²+2xy=9,故xy=[9-(x²+y²)]/2=(9-13)/2=-2
∴(y/x)+(x/y)=(x²+y²)/xy=-13/2