√[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+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|