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分式计算题: (x+2/x+1) % (x+3/x%2) % (x%4/x%3) +...

(x+2/x+1) - (x+3/x-2) - (x-4/x-3) + (x-5/x-4) =x+2/x+1 - x-3/x+2 - x+4/x+3 + x-5/x-4 =(x-x-x+x)+(2/x-3/x+4/x-5/x)+(1+2+3-4) =-2/x + 2

原式=[1+1/(x+1)]-[1+1/(x+3)]-[1+1/(x+2)]+[1+1/(x+4)] =1/(x+1)-1/(x+3)-1/(x+2)+1/(x+4) =[1/(x+1)+1/(x+4)]-[1/(x+2)+1/(x+3)] =[(2x+5)/(x+1)(x+4)]-[(2x+5)/(x+2)(x+3)] =(2x+5)[1/((x+1)(x+4)-1/(x+2)(x+3)] =(2x+5)[(x^2+5x+6)/(x+1)(x+...

解:3(x+2)=4(x+1) 3x+6=4x+4 4x-3x=6-4 x=2 希望能够帮助你

=[(x+2/x+1)-(x+3/x+2)]+[(x-5/x-4)-(x-4/x-3)] =[(x+2)(x+2)-(x+3)(x+1)]/(x+1)(x+2)+[(x-5)(x-3)-(x-4)(x-4)]/(x-4)(x-3) =1/(x+1)(x+2)+(-1)/(x-4)(x-3) =-10(x-1)/(x+1)(x+2)(x-3)(x-4) 顺便鄙视验证码

#include #include int main() { double x,num,sum=0; int i=1,j,flag=1; double jie; scanf("%lf",&x); while(1) { j=i; jie=1.0; while(j) {//计算阶乘 jie*=j--; } num=pow(x,i++)/jie;//pow为计算x的i次方 if(fabs(num)

X-2/3-x+3-1/X=3 1/X=-2/3 X=-3/2

(1)①x3y-xy3 =xy(x-+y)(x-y),②(x2+4)2-16x2=(x2+4-4x)(x2+4+4x),=(x-2)2(x+2)2;(2)①1x?3=x?43?x?2,1=-(x-4)-2(x-3),1=-x+4-2x+6,3x=9,x=3,当x=3时,x-3=0,x=3是增根,原方程无解;②x+4x?1-4x2?1=1,(x+4)(x+...

∵(x-3)(x+2)/(x-4)(x+2)=x-3/x-4 ∴x+2≠0 ∴x≠-2 ∵x-4为分母 ∴x-4≠0 ∴x≠4 要使分式x-3/x-4与(x-3)(x+2)/(x-4)(x+2)的值相等,则x应满足的条件是x≠-2且x≠4 如果本题有什么不明白可以追问,如果满意请点击“采纳为满意回答” 如果有其他问题请采纳本题...

﹛x-3≠0 x+2≠0 x-4≠0 ∴x的取值范围是x≠-2,且x≠3且x≠4的实数

(x+3)/(x+2)+(x+4)/(x+3)=(x+5)/(x+4)+(x+2)/(x+1) 1+1/(x+2)+1+1/(x+3)=1+1/(x+4)+1+1/(x+1) x+3+x+2/(x+2)(x+3)=x+1+x+4/(x+4)(x+1) 2x+5/(x+2)(x+3)=2x+5/(x+4)(x+1) (1)2x+5不为0 (x+2)(x+3)=(x+4)(x+1) x^2+5x+...

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