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

#include #include int main(void){double x,s,t;int i;while(scanf("%lf",&x)==1){for(s=0,i=1,t=x;t>1e-5;++i,t*=x/i)s+=i%2?t:-t;printf("%.2f\n",s);}return 0;}

解,得: 2-x/x-3=1/3-x -2 2-x==-1-2(x-3) 2-x==-1-2x+6 -x+2x==5-2 x==3 上面小明的错误在于两边都乘以(x-3)的时候,右边的-2忘记乘以(x-3)了,所以导致结果错误

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+...

1/(x-2)(x-3)-2/(x-1)(x-3)+1/(x-1)(x-2) =(x-1)/(x-2)(x-3)(x-1)-2(x-2)/(x-1)(x-2)(x-3)+(x-3)/(x-1)(x-2)(x-3) =【x-1-2x+4+x-3】/(x-1)(x-2)(x-3) =【0】/(x-1)(x-2)(x-3) =0

x-3/x-2+1=3/2-x 乘以(x-2)得 x-3+(x-2)=-3 x-3+x-2=-3 移项得 x+x=-3+3+2 合并同类项得 2x=2 x=1 经检验x=1是原方程的解

(1)x 2 -x+ 1 x 不是等式,故不是分式方程;(2) 1 a -3=a+4是分式方程;(3) 1 x -x=3 是无理方程,不是分式方程;(4) 20 x+y - 10 x-y =1是分式方程.故选B.

原方程化为: 1+1/(X+1)+1+1/(X+4)-1-1/(X+2)-1-1/(X+3)=0——化为真分式 1/(X+1)-1/(X+2)=1/(X+3)-1/(X+4) ——选择性移项 1/(X+1)(X+2)=1/(X+3)(X+4) ——两边同时通分, X^2+7X+12=X^2+3X+2 ——去分母, 4X=10 X=5/2 经检验: X=5/2是原方程的解。 ...

(x^2-x+1)/x=3化简得x+1/x=4,所以(x+1/x)^2=4^2,展开整理得到x^2+1/x^2=14 x^2/(x^4+x^2+1)分子和分母同时除以x^2得1/(x^2+1+1/x^2)=1/(1+14)=1/15

1。原式化为:2(x+1)=3x,2x+2=3x,x=2 2.原式化为:x(x-3)=(x-6)(x-2),x*2-3x=x*2-8x+12,x=2.4 3.原式化为:2x(5x-2)+5(2x+5)=(5x-2)(2x+5),x=7/3 4.原式化为:(x-1)(x-2)=(x-2),x=2,但x=2是增根舍去,方程没有根

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