(1/x-2 )+(2/x+1)-(2/x-1)-(1/x+2)
来源:百度知道 编辑:UC知道 时间:2024/06/09 10:22:08
1/(x-2 )+2/(x+1)-2/(x-1)-1/(x+2)
=[(x+2)-(x-2)]/(x+2)(x-2)+2[(x-1)-(x+1)]/(x+1)(x-1)
=4/(x^2-4)-4/(x^2-1)
=4[(x^2-1)-(x^2-4)]/(x^2-4)(x^2-1)
=12/(x^4-5x^2+4)
(1/x-2 )+(2/x+1)-(2/x-1)-(1/x+2)
=4/(X^2-4)-4/(X^2-1)
=12/(X^2-4)(X^2-1)
=1/X-2+2/X+1-2/X+1-1/X-2
=1/X+2/X-2/X-1/X-2+1+1-2
=-2
你的问题是[(1/x)-2 ]+[(2/x)+1]-[(2/x)-1]-[(1/x)+2] 还是[1/(x-2 )]+[2/(x+1)]-[2/(x-1)]-[1/(x+2)] 吖..我当作是[1/(x-2 )]+[2/(x+1)]-[2/(x-1)]-[1/(x+2)] 来做的..
(1/x-2 )+(2/x+1)-(2/x-1)-(1/x+2)
=4/(x^2-4)-4/(x^2-1)
=12/(x^2-4)(x^2-1)
1、4项通分计算得到 4/(x^2-4)
2、3项通分计算得到 -4/(x^2-1)
再把结果通分相加即得答案为 12/(x^2-4)(x^2-1)
通分 结果是3/(X+2)(X-2)(X+1)(X-1)
x/(x^2+x+1)=1/4 求x^2/(x^4+x^2+1)
x^2+x+1=2/(x^2+x)
[x+2]/[x+1]-[x+4]/[x+3]-[x+3]/[x+2]+]x+5]/[x+4]
((13 x-x^2)/(x+1)) (x+(13-x)/(x+1))=42
已知函数f(x)=a^x+(x-2)/(x+1) ,(a>1)
X=1/A-2+A,则√(4X+X*X)=????
1+x+x^2+x^3+x^4+x^5+……+x^2005(已知1+x+x^2+x^3+x^4=0)
1/2x — x+y 乘以 ( x+y/2x — x-y )
已知:F(X+1/X)=X^2/X^4+1, 求F(X)?
已知函数f(x)=(x^2+2x+a)/x,x∈[1,正无穷),