Question

求解这四道题！希望能把过程写在纸上发给我。

Answer 1

1、 比大小
这5个数可依次写成：(^后的数字表示指数)
2^(1/2), 2^(1/3), 2^(2/5), 2^(3/8), 2^(4/9)
指数通分：
2^(180/360), 2^(120/360), 2^(144/360), 2^(135/360), 2^(160/360)
所以从小到大的顺序是：
2^(120/360) < 2^(135/360) < 2^(144/360) < 2^(160/360) < 2^(180/360)
即 2^(1/3) < 2^(3/8) < 2^(2/5) < 2^(4/9) < 2^(1/2)

2、 化简
根号内的分子=2^30+2^20=2^20x(2^10+1)
根号内的分母=2^12+2^22=2^12x(1+2^10)
约分后，原式=根号内2^8
即原式=2^4=16

3、计算
原式=[(27/8)^(2/3) * (49/9)^(1/2) + (8/1000)^(2/3) ÷ (2/100)^(1/2) * (32/100)^(1/2)] ÷(625/10000)^(1/4)
=[9/4 * 7/3 + 1/25 * (100/2 * 32/100)^(1/2)] ÷5/10
=(21/4 +1/25 *4)*2
=(525/100 +16/100)*2
=541/100 *2
=541/50

4、 化简（为了便于你看懂，我把每个式子分开写）
(1). [a^(4/3) – 8a^(1/3)b]/[4b^(2/3) + 2a^(1/3)b^(1/3) + a^(2/3)]
= a^(1/3) * (a-8b)/[4b^(2/3) + 2a^(1/3)b^(1/3) + a^(2/3)]
= a^(1/3) * [a^(1/3)-2b^(1/3)][4b^(2/3) + 2a^(1/3)b^(1/3) + a^(2/3)]/[4b^(2/3) + 2a^(1/3)b^(1/3) + a^(2/3)] (注：这里是一个立方差公式)
= a^(1/3) * [a^(1/3)-2b^(1/3)]
(2). ÷[a^(-2/3) – 2b^(1/3)/a]
= a/[a^(1/3) -2b^(1/3)]
(3). [a*a^(2/3)]^(1/2) / [a^(1/2) * a^(1/3)]^(1/5)
= a^(5/6) / a^(1/6)
= a^(2/3)
所以原式= a^(1/3) * [a^(1/3)-2b^(1/3)] * a/[a^(1/3) -2b^(1/3)] * a^(2/3) = a²

Answer 2