2012贵州先化简1/x+1-3-x/x2-6x+91/(x+1)-[(3-x)/(x^2-6x+9)]÷[(x^2+x)/(x-3)]=1/(x+1)+[(x-3)/(x-3)^2][(x-3)/(x^2+x)]=1/(x+1)+1/(x^2+x)=1/(x+1
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 14:38:25
![2012贵州先化简1/x+1-3-x/x2-6x+91/(x+1)-[(3-x)/(x^2-6x+9)]÷[(x^2+x)/(x-3)]=1/(x+1)+[(x-3)/(x-3)^2][(x-3)/(x^2+x)]=1/(x+1)+1/(x^2+x)=1/(x+1](/uploads/image/z/3930790-22-0.jpg?t=2012%E8%B4%B5%E5%B7%9E%E5%85%88%E5%8C%96%E7%AE%801%2Fx%2B1-3-x%2Fx2-6x%2B91%2F%EF%BC%88x%EF%BC%8B1%EF%BC%89%EF%BC%8D%EF%BC%BB%EF%BC%883%EF%BC%8Dx%EF%BC%89%2F%EF%BC%88x%5E2%EF%BC%8D6x%EF%BC%8B9%EF%BC%89%EF%BC%BD%C3%B7%EF%BC%BB%EF%BC%88x%5E2%EF%BC%8Bx%EF%BC%89%2F%EF%BC%88x%EF%BC%8D3%EF%BC%89%EF%BC%BD%EF%BC%9D1%2F%EF%BC%88x%EF%BC%8B1%EF%BC%89%EF%BC%8B%EF%BC%BB%EF%BC%88x%EF%BC%8D3%EF%BC%89%2F%EF%BC%88x%EF%BC%8D3%EF%BC%89%5E2%EF%BC%BD%EF%BC%BB%EF%BC%88x%EF%BC%8D3%EF%BC%89%2F%EF%BC%88x%5E2%EF%BC%8Bx%EF%BC%89%EF%BC%BD%EF%BC%9D1%2F%EF%BC%88x%EF%BC%8B1%EF%BC%89%EF%BC%8B1%2F%EF%BC%88x%5E2%EF%BC%8Bx%EF%BC%89%EF%BC%9D1%2F%EF%BC%88x%EF%BC%8B1)
2012贵州先化简1/x+1-3-x/x2-6x+91/(x+1)-[(3-x)/(x^2-6x+9)]÷[(x^2+x)/(x-3)]=1/(x+1)+[(x-3)/(x-3)^2][(x-3)/(x^2+x)]=1/(x+1)+1/(x^2+x)=1/(x+1
2012贵州先化简1/x+1-3-x/x2-6x+9
1/(x+1)-[(3-x)/(x^2-6x+9)]÷[(x^2+x)/(x-3)]
=1/(x+1)+[(x-3)/(x-3)^2][(x-3)/(x^2+x)]
=1/(x+1)+1/(x^2+x)
=1/(x+1)+1/[x(x+1)]
=(x+1)/[x(x+1)]
=1/x
=1/√2
=√2/2
为什么还要 =√2/2
2012贵州先化简1/x+1-3-x/x2-6x+91/(x+1)-[(3-x)/(x^2-6x+9)]÷[(x^2+x)/(x-3)]=1/(x+1)+[(x-3)/(x-3)^2][(x-3)/(x^2+x)]=1/(x+1)+1/(x^2+x)=1/(x+1
一般分母不能是无理数形式,这是规定
您好,很高兴为您解答,skyhunter002为您答疑解惑
如果本题有什么不明白可以追问,如果满意记得采纳
如果有其他问题请采纳本题后另发点击向我求助,答题不易,请谅解,谢谢.
祝学习进步