
2个回答
展开全部
(n+1)²+n²+n²(n+1)²
=(n+1)²-2n(n+1)+n²+2n(n+1)+n²(n+1)²
=[(n+1)-n]²+2n(n+1)+n²(n+1)²
=1+2n(n+1)+n²(n+1)²
=(1+n+n²)²
通分
原式=√{[(n+1)²+n²+n²(n+1)²]/n²(n+1)²}
=√[(1+n+n²)²/n²(n+1)²]
=(1+n+n²)/(n+n²)
=(n+1)²-2n(n+1)+n²+2n(n+1)+n²(n+1)²
=[(n+1)-n]²+2n(n+1)+n²(n+1)²
=1+2n(n+1)+n²(n+1)²
=(1+n+n²)²
通分
原式=√{[(n+1)²+n²+n²(n+1)²]/n²(n+1)²}
=√[(1+n+n²)²/n²(n+1)²]
=(1+n+n²)/(n+n²)
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