设S=1+根号2分之1+根号2+根号3分之1+根号3+根号4分之1+.+根号2003+根号2004分之1,t=1-2+3-4+5-6+.++2003-2004,求(S+1)的平方分之t的值
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![设S=1+根号2分之1+根号2+根号3分之1+根号3+根号4分之1+.+根号2003+根号2004分之1,t=1-2+3-4+5-6+.++2003-2004,求(S+1)的平方分之t的值](/uploads/image/z/2640252-12-2.jpg?t=%E8%AE%BES%3D1%2B%E6%A0%B9%E5%8F%B72%E5%88%86%E4%B9%8B1%2B%E6%A0%B9%E5%8F%B72%2B%E6%A0%B9%E5%8F%B73%E5%88%86%E4%B9%8B1%2B%E6%A0%B9%E5%8F%B73%2B%E6%A0%B9%E5%8F%B74%E5%88%86%E4%B9%8B1%2B.%2B%E6%A0%B9%E5%8F%B72003%2B%E6%A0%B9%E5%8F%B72004%E5%88%86%E4%B9%8B1%2Ct%3D1-2%2B3-4%2B5-6%2B.%2B%2B2003-2004%2C%E6%B1%82%EF%BC%88S%2B1%29%E7%9A%84%E5%B9%B3%E6%96%B9%E5%88%86%E4%B9%8Bt%E7%9A%84%E5%80%BC)
设S=1+根号2分之1+根号2+根号3分之1+根号3+根号4分之1+.+根号2003+根号2004分之1,t=1-2+3-4+5-6+.++2003-2004,求(S+1)的平方分之t的值
设S=1+根号2分之1+根号2+根号3分之1+根号3+根号4分之1+.+根号2003+根号2004分之1,t=1-2+3-4+5-6+.+
+2003-2004,求(S+1)的平方分之t的值
设S=1+根号2分之1+根号2+根号3分之1+根号3+根号4分之1+.+根号2003+根号2004分之1,t=1-2+3-4+5-6+.++2003-2004,求(S+1)的平方分之t的值
考察平方根分式的化简:S=1/.(√1+√2)+1/(√2+√3)+...+1/(√2003+√2004)=√2-√1+√3√-√2+√4-√3+...+√2003-√2002+√2004-√2003,逐项抵消,最终S=√2004-√1,t=(1-2)+(3-4)+(5-6)+...+(2003-2004)=-1x(2004/2)=-1002.则所求值为t/(S+1)²=(-1002)/2004=-1/2.
t=1-2+3-4+5-6+....+2003-2004
=-1-1-1-........-1
=-1*1002
=-1002
s=1/(1+√2)+1/(√2+√3)+1/(√3+√4)+........+1/(√2003-√2004)
=(√2-1)/[(1+√2)(√2-1)]+(√3-√2)/[(√2+√3)(√3-√2)]+(√4-√3)/[(√3...
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t=1-2+3-4+5-6+....+2003-2004
=-1-1-1-........-1
=-1*1002
=-1002
s=1/(1+√2)+1/(√2+√3)+1/(√3+√4)+........+1/(√2003-√2004)
=(√2-1)/[(1+√2)(√2-1)]+(√3-√2)/[(√2+√3)(√3-√2)]+(√4-√3)/[(√3+√4)(√4-√3)]+........+(√2004-√2003)/[(√2003+√2004)(√2004-√2003)]
=√2-1+√3-√2+√4-√3+........+√2004-√2003
=√2004-1
t/(s+1)^2
=-1002/(√2004-1+1)^2
=-1002/2004
=-1/2
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