Question

高数 求定积分 划线处怎么算啊

Answer 1

解答

1.
(a-b)(a2+ab+b2)=a3-b3
所以原式=(5x)3-(1/2y)3=125x3-1/8×y3
2.因为不清楚你的√ab是指√a×b,还是√ab
但是可设√a=x,√b=y 然后进行通分,并运用公式运算
3.应该没得化简了吧, 如果(1)为分子(2)为分母那就可化简
分子分母同乘以分子后 得
=(4+2√3)/√[(4-√12)×(4-2√3)]
=(4+2√3)/√[42-(2√3)2]
=(4+2√3)/√4
=(4+2√3)/2
=2+√3
4.
原式=（2+31/2)1/2 ×{[2+(2+31/2)1/2]×[(2-(2+31/2)1/2]}1/2
=（2+31/2)1/2 ×{22-[(2+31/2)1/2]2}1/2
=（2+31/2)1/2 ×[4-(2+31/2)]1/2
=（2+31/2)1/2 ×4-（2+31/2)1/2 ×(2+31/2)1/2
=（2+31/2)1/2 ×4-(2+31/2)
5.
原式=[2xy(x2-y2)+(x2-y2)(x-y)2]2 ÷(x2+y2)2
=[(x2-y2)(2xy+x2-2xy+y^)]2 ÷(x2+y2)2
=(x2-y2)2 * (x2+y2)2 ÷(x2+y2)2
=(x2-y2)2
6.
a2x=√2+1
a-2x=1/(√2 +1 )=√2 -1
设ax=x,a-x=y
则x=√(√2 +1 ),y=√(√2-1)
则原式=(x3+y3)/(x+y)=x2-xy+y2
=(√2+1)-√[(√2+1)(√2-1)]+(√2-1)
=2√2-√(2-1)
=2√2-1
7.
直接求出a 再代入求值
8.通分后得
=[1/(x+1) -2- 1/(x-1)]*(x2-1)/[x(x2-1)] + x2/[(x2-1)×x]
=[(x-1)-2(x2-1)-(x+1)]/[x(x2-1)] + x2/[x(x2-1)]
=-2x2/[x(x2-1)] + x2/[x(x2-1)]
= -x2/[x(x2-1)]
=-x/(x2-1)
看了一下楼下的,原来题3是这样化解的.
题7都他的方法为好!
另题5是可以化解的,用的公式是 (a+b)(a-b)=a2-b2

Answer 2

如图

Answer 3

希望写的很清楚