a²+2ab+b²=8,a²-b²=2倍根号2,则a-b分之b等于多少?

a²+2ab+b²=8,a²-b²=2倍根号2,则a-b分之b等于多少?
a²+2ab+b²=8,a²-b²=2倍根号2,则a-b分之b等于多少?

a²+2ab+b²=8,a²-b²=2倍根号2,则a-b分之b等于多少?
(a+b)²=8
a+b=±2√2
a²-b²=(a+b)(a-b)=2√2
所以
a+b=-2√2,a-b=-1
或a+b=2√2,a-b=1
相减
a+b=-2√2,a-b=-1,b=(1-2√2)/2
a+b=2√2,a-b=1,b=(2√2-1)/2
所以原式=(2√2-1)/2

2倍根号2除以根号8

a²+2ab+b²=8
(a+b)²=8 (1)
a²-b²=2√2
(a+b)(a-b)=2√2 (2)
(1)/(2)
(a+b)/(a-b)
=8/2√2
(a+b)/(a-b)、
=2√2
由此，得
(2√2-1)a=(2√...

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a²+2ab+b²=8
(a+b)²=8 (1)
a²-b²=2√2
(a+b)(a-b)=2√2 (2)
(1)/(2)
(a+b)/(a-b)
=8/2√2
(a+b)/(a-b)、
=2√2
由此，得
(2√2-1)a=(2√2+1)b
a=(2√2+1)b/(2√2-1)=(2√2+1)²b/7=(9+4√2)b/7
b/(a-b)=b/[(9+4√2)b/7 -b]
=7/(9+4√2-7)
=7/(4√2+2)
=7(4√2-2)/28
=(2√2-1)/2

收起

a²+2ab+b²=8 → (a+b)²=8 ①
a²-b²=2√2 → (a+b)(a-b)=2√2 ②


b/(a-b)=((a+b)/(a-b)-1)/2=(a+b)²/(2*(a+b)(a-b))-1/2=√2-1/2