数学n为正整数,化解根号【1+1
应该是√[1+1/n²+1/(n+1)²]吧?
解:
1+1/n²+1/(n+1)²
={[n(n+1)]²+(n+1)²+n²}/[n(n+1)]²
={(n²+1)*(n+1)²+n²}/[n(n+1)]²
={(n²+1)*(n²+2n+1)+n²}/[n(n+1)]²
令n²+1=a
原式可化为
={a*(a+2n)+n²}/n(n+1)]²
=(a²+2an+n²)/[n(...全部
应该是√[1+1/n²+1/(n+1)²]吧?
解:
1+1/n²+1/(n+1)²
={[n(n+1)]²+(n+1)²+n²}/[n(n+1)]²
={(n²+1)*(n+1)²+n²}/[n(n+1)]²
={(n²+1)*(n²+2n+1)+n²}/[n(n+1)]²
令n²+1=a
原式可化为
={a*(a+2n)+n²}/n(n+1)]²
=(a²+2an+n²)/[n(n+1)]²
=(a+n)²/[n(n+1)]²
所以√[1+1/n²+1/(n+1)²]=(a+n)/n(n+1)=(n²+n+1)/n(n+1)
另一形式
1+1/n²+1/(n+1)²
=1+(n²+n²+2n+1)/n²(n+1)²
=1+2(n²+n)/n²(n+1)²+1/n²(n+1)²
=1+2/n(n+1)+1/n²(n+1)²
=[1+1/n(n+1)]²
故原式=1+1/n(n+1)
。
收起