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123456789x987654321=?

123456789*987654321-123456788*987654322=(123456788+1)*987654321-123456788*(987654321+1) =987654321-123456788 =864197533

A=987654321*(12345678+1)=987654321*12345678+987654321 B= (987654321+1)*12345678=987654321*12345678+12345678 因为987654321>12345678 所以A>B 满意请采纳

经我用计算机一算得:123456789乘以987654321=121932631112635269 报告:答案完全正确,哈哈

第一个等于1.219326302乘10的17次方

设123456788=a,987654321=b A=(a+1)b=ab+b B=a(b+1)=ab+a 因为a〈b,所以A〉B

另a=987654321,b=123456789, 那么A = a×b,B=(a+1)(b-1)=a×b+b-a-1=a×b+123456789-987654321-1 所以A-B>0,A>B

/求两个数的最大公约数和最小公倍数#include "stdafx.h"#includeint gongyueshu(int x,int y);int gongbeishu(int x,int y);using namespace std;int main(int argc, char* argv[]){ int z,p; z=gongyueshu(9,6); p=gongbeishu(6,9); cout

(123456788+1)x987654322〉123456788x987654321

123456789*987654321-123456788*987654322 =123456789×987654321-(123456789-1)×(987654321+1) =123456789×987654321-(123456789×987654321-987654321+123456789-1) =123456789×987654321-123456789×987654321+987654321-123...

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