n的平方+3n=1,求n(n+1)(n+2)+1的值.(初一因式分解内容)谢谢!
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n的平方+3n=1,求n(n+1)(n+2)+1的值.(初一因式分解内容)谢谢!
n的平方+3n=1,求n(n+1)(n+2)+1的值.(初一因式分解内容)谢谢!
n的平方+3n=1,求n(n+1)(n+2)+1的值.(初一因式分解内容)谢谢!
把 n(n+1)(n+2)+1 打开、得n的平方+3n+2+1
因为 n的平方+3n=1
所以 n的平方+3n+2+1=1+2+1
=4
由n^2+3n=1得
n(n+1)(n+2)+1=n(n^2+3n+2)=3n+1
由n^2+3n=1得
n(n+1)(n+2)+1 =n(n^2+3n+2)+1 =n(1+2)+1 =3n+1 设y=3n+1 n^2+3n=1 9n^2+27n=9 9n^2+6n+1+21n=9+1 (3n+1)^2+21n+7=10+7 (3n+1)^2+7(3n+1)=17 y^2+7y+49/4=17+49/4 (y+7/2)^2=117/4