√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2√[1+1/(2^2)+1/(3^2)] = 1+1/2-1/3=1+1/6√[1+1/(3^2)+1/(4^2)] = 1+1/3-1/4=1+1/12根据上述等式揭示的规律,写出用字母n (n为大于1)的自然数表示这一规律的等式,并说明等式成立的

来源：学生作业帮助网 编辑：作业帮 时间：2024/07/13 11:06:00
√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2√[1+1/(2^2)+1/(3^2)] = 1+1/2-1/3=1+1/6√[1+1/(3^2)+1/(4^2)] = 1+1/3-1/4=1+1/12根据上述等式揭示的规律,写出用字母n (n为大于1)的自然数表示这一规律的等式,并说明等式成立的
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√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2√[1+1/(2^2)+1/(3^2)] = 1+1/2-1/3=1+1/6√[1+1/(3^2)+1/(4^2)] = 1+1/3-1/4=1+1/12根据上述等式揭示的规律,写出用字母n (n为大于1)的自然数表示这一规律的等式,并说明等式成立的
√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2
√[1+1/(2^2)+1/(3^2)] = 1+1/2-1/3=1+1/6
√[1+1/(3^2)+1/(4^2)] = 1+1/3-1/4=1+1/12
根据上述等式揭示的规律,写出用字母n (n为大于1)的自然数表示这一规律的等式,并说明等式成立的理由

√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2√[1+1/(2^2)+1/(3^2)] = 1+1/2-1/3=1+1/6√[1+1/(3^2)+1/(4^2)] = 1+1/3-1/4=1+1/12根据上述等式揭示的规律,写出用字母n (n为大于1)的自然数表示这一规律的等式,并说明等式成立的
规律：√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))
证明：假设√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))成立
则：（1+1/(n+1)-1/(n+2))^2=1+2/(n+1)-2/(n+2)+1/((n+1)^2)+1/((n+2)^2)-2/((n+1)(n+2))
=1+1/((n+1)^2)+1/((n+1)^2)
即：1+1/((n+1)^2)+1/((n+1)^2)=（1+1/(n+1)-1/(n+2))^2
两边各开根号：得：√1+1/((n+1)^2)+1/((n+1)^2)=1+1/(n+1)-1/(n+2)
又n (n为大于1)的自然数,n=1时等式成立
综上所述：√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))

1/√(1+2) (√1-x)^2 √2x+1 2/√3-1 (√2+1-2分之1√2)^2 (√2 1-2分之1√2)^2 化简1/1+√2+1/√2+√3+...+1/√8+√9 1/1+√2+1/√2+√3+…+1/√2013+√2014= 实数的运算|1-√2|*1/(1-√2) √2+1,√2-1,1能否构成直角三角形 √2+1是怎样变成 1/√2-1 1/(1+√2)+1/(√2+√3)+1/(√3+√4)+.1/(√2012+√2013) 1/(√2009+√2008)+1/(√2008+√2007)+.+1/(√3+√2)+1/(√2+1) √(a-1)(a^2-1) （-1 化简√(1-√3/2)/2 √（2^2-1)=√3, 2+√2-1的绝对值-（√2+1） |√6-√2|+|1-√2|+|√6-3|