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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+...

=[(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

(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+...

解: [(x+1)+1]/(x+1)-[(x+3)+1]/(x+3)=[(x+5)+1]/(x+5)-[(x+7)+1]/(x+7) (x+1)/(x+1)+1/(x+1)-(x+3)/(x+3)-1/(x+3)=(x+5)/(x+5)+1/(x+5)-(x+7)/(x+7)-1/(x+7) 1+1/(x+1)-1-1/(x+3)=1+1/(x+5)-1-1/(x+7) 1/(x+1)-1/(x+3)=1/(x+5)-1/(x+7) [(x+3)...

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

#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)

解: (1) 4/(8-x)>0 8-x>0 x

第一题,依次计算,(int)(x+y)得7,a%3得1,1*7得7,7%2得1,1%4得0,x+0得2.5。

(1)x+1分之x+2 -x+2分之x+3=x+3分之x+4 -x+4分之x+5 x+1分之x+2+x+4分之x+5= x+2分之x+3+x+3分之x+4 [(x+4)(x+2)+(x+5)(x+1)]/(x+1)(x+4)=[(x+3)(x+3)+(x+4)(x+2)]/(x+3)(x+2) (x²+6x+8+x²+6x+5)/(x+1)(x+4)=(x²+6x+9+x²+...

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