
因式分解的问题
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(m+1)(n+1)
(m+1)(n+1)(p+1)
(m+1)(n+1)(p+1)
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1.因式分解1+m+n+mn
=(1+m)+(n+mn)
=(1+m)+n(1+m)
=(1+m)(1+n)
2.由此,因式分解
1+m+n+p+mn+mp+np+mnp
=(1+m)+(n+p)+(mn+mp)+(np+mnp)
=(1+m)+(n+p)+m(n+p)+np(1+m)
=[(1+m)+np(1+m)]+[(n+p)+m(n+p)]
=(1+m)(1+np)+(n+p)(1+m)
=(1+m)(1+np+n+p)
=(1+m)[(1+n)+(np+p)]
=(1+m)[(1+n)+p(1+n)]
=(1+m)(1+n)(1+p)
=(1+m)+(n+mn)
=(1+m)+n(1+m)
=(1+m)(1+n)
2.由此,因式分解
1+m+n+p+mn+mp+np+mnp
=(1+m)+(n+p)+(mn+mp)+(np+mnp)
=(1+m)+(n+p)+m(n+p)+np(1+m)
=[(1+m)+np(1+m)]+[(n+p)+m(n+p)]
=(1+m)(1+np)+(n+p)(1+m)
=(1+m)(1+np+n+p)
=(1+m)[(1+n)+(np+p)]
=(1+m)[(1+n)+p(1+n)]
=(1+m)(1+n)(1+p)
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