2 mathematics questions

2006-11-27 12:15 am
1. please make m be the subject of the formula where m and n are positive.

m+n/2m+n = m/n

the answer is m= n/(平方根2)


2. 2b/(a-b)(a+b) + 2a/a^2+ab + b^2/b-a = ?

please show the steps

thanks!!
更新1:

the answer of question 2 is 2a-ab^2-b^3/(a-b)(a+b)

回答 (1)

2006-11-27 12:35 am
✔ 最佳答案
我加上了括號使題目更易了解,希望我理解沒錯。

1. please make m be the subject of the formula where m and n are positive.

(m+n)/(2m+n) = m/n

n(m+n) = m(2m+n)

mn + n² = 2m² + mn

2m² = n²

m² = n²/2

m = √(n²/2)

m = n/√2

========================================
2. 2b/(a-b)(a+b) + 2a/a²+ab + b²/b-a = ?

2b/(a-b)(a+b) + 2a/(a²+ab) + b²/(b-a)

= 2b/(a-b)(a+b) + 2a/[a(a+b)] - b²/(a-b)

= 2b/(a-b)(a+b) + 2/(a+b) - b²/(a-b)

= 2b/(a+b)(a-b) + 2(a-b)/[(a+b)(a-b)] - b²(a+b)/[(a+b)(a-b)]

= [2b + 2(a-b) - b²(a+b)] / [(a+b)(a-b)]

= [2b + 2a - 2b - b²(a+b)] / [(a+b)(a-b)]

= (2a - ab² - b³) / [(a+b)(a-b)]

2006-11-26 16:38:19 補充:
小小補充:第一條題目中,因為 m 和 n 都不是 0,所以應用了交差相乘。第二條題目中,首先將三項都通份母為 (a+b)(a-b),再將分子展開化簡,就能得到答案。希望幫到你。


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