Question

(123456789/123456788-123456788/123456789)÷(1/123456788+1/123456789)

简便算法

Answer 1

转换一下，a=123456788，b=123456789
原式=(b/a-a/b)/(1/a+1/b)
=[(b^2-a^2)/ab]/[(b+a)/ab]
=(b^2-a^2)/(b+a)
=(b+a)(b-a)/(b+a)
=b-a
=123456789-123456788
=1

Answer 2

X=123456789，123456788=X-1
原式=[X/（X-1）-（X-1）/X]/[1/（X-1）+1/X]
=[（X²-（X-1）²）/（X（X-1））]/[（X+（X-1））/（X（X-1））]
=[X²-（X²-2X+1）]/[2X-1]
=（2X-1）/（2X-1）
= 1
答案与X无关

Answer 3

设 123456789 = a, 123456788 = b；
则原式变作
( a/b - b/a ) / ( 1/b + 1/a ) = ( a^2 - b^2) / ab * ab / ( a + b) = a - b = 123456789 - 123456788 = 1;

答案是1，
正确的话 给个最佳答案，码字不容易呢

Answer 4

原式=(1 1/123456788-1 1/123456789)/(1/123456788 1/123456789)=1
简便算法的核心就是把分子分离一下化简 如123456789/123456788=(123456788 1)/123456788

Answer 5

(123456789/123456788-123456788/123456789)÷(1/123456788+1/123456789)
=(1+1/123456788-1+1/123456789)÷(1/123456788+1/123456789)
=(1/123456788+1/123456789)÷(1/123456788+1/123456789)
=1

Answer 6

(123456789/123456788-123456788/123456789)÷(1/123456788+1/123456789)

=[(123456789^2-123456788^2)/(123456788*123456789)]/[(123456788+123456789)/(123456788*123456789)]

=(123456789^2-123456788^2)/(123456788+123456789)

=123456789-123456788=1