分解因式:x(x+1)(x+2)(x+3)+1
5个回答
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原式=(x²+3x+2)(x²+3x)+1
令x²+3x为a
=a(a+2)+1
=a²+2a+1
=(x²+3x)²+2(x²+3x)+1
=x四次方+6x三次方+11X²+6x+1
令x²+3x为a
=a(a+2)+1
=a²+2a+1
=(x²+3x)²+2(x²+3x)+1
=x四次方+6x三次方+11X²+6x+1
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x(x+1)(x+2)(x+3)+1
=[x(x+3)][(x+1)(x+2)]+1
=(x^2+3x)(x^2+3x+2)+1
设 x^2+3x =y
=y(y+2)+1
=y^2+2y+1
=(y+1)^2
=(x^2+3x+1)^2
=[x(x+3)][(x+1)(x+2)]+1
=(x^2+3x)(x^2+3x+2)+1
设 x^2+3x =y
=y(y+2)+1
=y^2+2y+1
=(y+1)^2
=(x^2+3x+1)^2
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x(x+1)(x+2)(x+3)+1
=x(x+3)(x+1)(x+2)+1
=(x*x+3x+1-1)(x*x+3x+1+1)+1
=(x*x+3x+1)^2 -1+1
=(x*x+3x+1)^2
=x(x+3)(x+1)(x+2)+1
=(x*x+3x+1-1)(x*x+3x+1+1)+1
=(x*x+3x+1)^2 -1+1
=(x*x+3x+1)^2
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展开全部
x(x+1)(x+2)(x+3)+1
=(x^2+3x)(x^2+3x+2)+1
=(x^2+3x)^2+2(x^2+3x)+1
=(x^2+3x+1)^2
=(x^2+3x)(x^2+3x+2)+1
=(x^2+3x)^2+2(x^2+3x)+1
=(x^2+3x+1)^2
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