Question

八年级上册数学因式分解

Answer 1

解(1) 16a^4-b^4 ^表示乘方
=(4a²)²-(b²)²
=(4a²+b²)(4a²-b²)
=(4a²+b²)(2a+b)(2a-b)

解(2) a²(m-n)+b²(n-m)
=a²(m-n)-b²(m-n)
=(m-n)(a²-b²)
=(m-n)(a+b)(a-b)

解(3) 3(a+1)²-75(a-4)²
=3[(a+1)²-25(a-4)²]
=3﹛(a+1)²-[5(a-4)]²﹜
=3[(a+1)+5(a-4)][(a+1)-5(a-4)]
=3(a+1+5a-20)(a+1-5a+20)
=3(6a-19)(-4a+21)
=-3(6a-19)(4a-21)

解(4) a^(2n+1)-16a^(2n-1)
=[a^(2n-1)]×(a²-16)
=[a^(2n-1)](a+4)(a-4)

解(5) 9(a-b)²-25(a+b)²
=[3(a-b)]²-[5(a+b)]²
=[3(a-b)+5(a+b)][(3(a-b)-5(a+b)]
=(3a-3b+5a+5b)(3a-3b-5a-5b)
=(8a+2b)(-2a-8b)
= -4(4a+b)(a+4b)

Answer 2

引用自由的数学鸟的回答：
解(1) 16a^4-b^4 ^表示乘方
=(4a²)²-(b²)²
=(4a²+b²)(4a²-b²)
=(4a²+b²)(2a+b)(2a-b)

解(2) a²(m-n)+b²(n-m)
=a²(m-n)-b²(m-n)
=(m-n)(a²-b²)
=(m-n)(a+b)(a-b)

解(3) 3(a+1)²-75(a-4)²
=3[(a+1)²-25(a-4)²]
=3﹛(a+1)²-[5(a-4)]²﹜
=3[(a+1)+5(a-4)][(a+1)-5(a-4)]
=3(a+1+5a-20)(a+1-5a+20)
=3(6a-19)(-4a+21)
=-3(6a-19)(4a-21)

解(4) a^(2n+1)-16a^(2n-1)
=[a^(2n-1)]×(a²-16)
=[a^(2n-1)](a+4)(a-4)

解(5) 9(a-b)²-25(a+b)²
=[3(a-b)]²-[5(a+b)]²
=[3(a-b)+5(a+b)][(3(a-b)-5(a+b)]
=(3a-3b+5a+5b)(3a-3b-5a-5b)
=(8a+2b)(-2a-8b)
= -4(4a+b)(a+4b)

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