1、(x+1)(x-1)(x²+1)…(x^32+1)
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1、(x+1)(x-1)(x²+1)…(x^32+1)
2、已知a²+b²=1,a-b=1/2,求a²b²、(a+b)^4
2、已知a²+b²=1,a-b=1/2,求a²b²、(a+b)^4
1、(x+1)(x-1)(x²+1)…(x^32+1)
=(x²-1)(x²+1)…(x^32+1)
=(x^4-1)(x^4+1)…(x^32+1)
=(x^8-1)(x^8+1)…(x^32+1)
=(x^16-1)(x^16+1)(x^32+1)
=(x^32-1)(x^32+1)
=x^64-1
2、(a-b)²=a²+b²-2ab
ab=[(a-b)²-(a²+b²)]/2=(1/4-1)/2=3/8
a²b²=(ab)²=9/64
(a+b)^4=[(a+b)²]²=[(a²+b²+2ab)]²=[1+2*3/8]²=(7/4)²=49/16
=(x²-1)(x²+1)…(x^32+1)
=(x^4-1)(x^4+1)…(x^32+1)
=(x^8-1)(x^8+1)…(x^32+1)
=(x^16-1)(x^16+1)(x^32+1)
=(x^32-1)(x^32+1)
=x^64-1
2、(a-b)²=a²+b²-2ab
ab=[(a-b)²-(a²+b²)]/2=(1/4-1)/2=3/8
a²b²=(ab)²=9/64
(a+b)^4=[(a+b)²]²=[(a²+b²+2ab)]²=[1+2*3/8]²=(7/4)²=49/16
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