(√2-1)^2012·(√2+1)^2013=
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(√2-1)^2012·(√2+1)^2013=
![(√2-1)^2012·(√2+1)^2013=](/uploads/image/z/15793696-64-6.jpg?t=%EF%BC%88%E2%88%9A2-1%EF%BC%89%5E2012%C2%B7%EF%BC%88%E2%88%9A2%2B1%EF%BC%89%5E2013%3D)
(√2-1)^2012·(√2+1)^2013
=(√2-1)^2012·(√2+1)^2012·(√2+1)
=【(√2-1)·(√2+1)】^2012·(√2+1)
=1^2012·(√2+1)
=√2+1
=(√2-1)^2012·(√2+1)^2012·(√2+1)
=【(√2-1)·(√2+1)】^2012·(√2+1)
=1^2012·(√2+1)
=√2+1
二次根式题(1/√2+1+1/√3+√2+1/√4+√3+.+1/√2013+2012)(√2013 +1)= ps:+
请计算1/(2√1+1√2)+1/(3√2+2√3)+∧+1/[(2013√2012)+(2012√2013)]
计算:√(1-√2)²+√(√2-√3)²+√(√3-√4)²+···+√(√2004-√
(√2 +1分之1+√3+√2分之1+.+√2012+√2011分之1)(√2012+1)=?
1/(1+√2)+1/(√2+√3)+···+1/(√5+√6)=...如题
1/(1+√2)+1/(√2+√3)+1/(√3+√4)+.1/(√2012+√2013)
(√3+√2)*(√3-√2)= √1/7+√28-√700=
证明:2(√(n+1)-1)<1+1/√2+1/√3+···+1/√n<2√n
(√6+1)2014-2(√6+1)2013-5(√6+1)2012+2013
√1/5=(),计算:√2/3-√3/2,√18-√8+√1/8,√25/2+√99-√18
实数的运算 [急]√96-√24=√3(√3-1)=√20+√5/√5-√1/3*√2=√54=√2-5√1/2+√32
m=√(7+2√12);m=1/(2+√2);m=√(2-√3)+√(2+√3).