java try catch:化简:x(x+1)\1+(x+1)(x+2)\1+(x+2)(x+3)\1+(x+3)(x+4)\1
来源:百度文库 编辑:高校问答 时间:2024/09/21 09:26:31
应该系化简:化简:x(x+1)/1+(x+1)(x+2)/1+(x+2)(x+3)/1+(x+3)(x+4)/1
如果1是分子,就应该这么算
原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/x-1/(x+4)
如果1是分母,这题就没意思了,把括号都打开,合并同类项即可
1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/x-1/(x+4)=3/(x(x-4))
x(x+1)\1+(x+1)(x+2)\1+(x+2)(x+3)\1+(x+3)(x+4)\1
=x\1-(x+1)\1+.........(x+3)\1-(x+4)\1
=x\1-(x+4)\1
=x(x+4)\4
/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/x-1/(x+4)=3/(x(x-4))
x(x+1)\1+(x+1)(x+2)\1+(x+2)(x+3)\1+(x+3)(x+4)\1
=x\1-(x+1)\1+.........(x+3)\1-(x+4)\1
=x\1-(x+4)\1
=x(x+4)\4
(x*x*x-x*x+x)/(x*x*x+1)+(x*x+x+1)/[(1-x*x)/x)]
(x*x*x-x*x+x)/(x*x*x+1)+(x*x+x+1)/[(1-x*x)/x)]
化简:x(x+1)\1+(x+1)(x+2)\1+(x+2)(x+3)\1+(x+3)(x+4)\1
1+x+x(1+x)+x(1+x)2+x(1+x)3
1+x+x(1+x)+x(1+x)2+x(1+x)3
已知x*x-3x+1=0,求x*x+1/x*x
函数f(x)=x*x*x-x*x-x+1有多少个零点?
1+x+x^2+x^3=0 ,求x+x^2+x^3+...+x^2000
X*X-2X-1=0 求2x*x*x-3*x*x-4*x+2
x*(1+4x*x){0.5}积分?