设m-n=1⼀4,m+n=2,求[m눀n눀⼀(m눀+2mn+n눀)-2⼀mn÷(1⼀m+1⼀n)눀]●1⼀(m-n)的值

2025-05-24 01:21:52
推荐回答(3个)
回答1:

m-n=1/4 ......(1)
m+n=2 ......(2)
【(2)²-(1)²】÷2得:mn = 2²-(1/4)²=63/16 ......(3)

[ m²n²/(m²+2mn+n²) - 2/(mn)÷(1/m+1/n)² ] × 1/(m-n)
= [ m²n²/(m+n)² - 2/(mn)÷(m+n)²/(m²n²) ] × 1/(m-n)
= [ m²n²/(m+n)² - 2/(mn)×(m²n²)/(m+n)² ] × 1/(m-n)
= [ m²n²/(m+n)² - 2(mn)/(m+n)² ] × 1/(m-n)
= mn(mn-2)/(m+n)² × 1/(m-n)
将(1)、(2)、(3)代入上式:
= 63/16 ×(63/16-2) /2² × 1/(1/4)
= 1953/256

回答2:

把条件联立解方程组

回答3:

m=9/8 n=7/8 算式最后化解为(2m^2n^2-mn)/8=(81*49-63*4)/(32*64)