关于分式计算的题2/(x+1)(x+3)+2/(x+3)(x+5)+…+2/(x+99)(x+101)=?
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关于分式计算的题
2/(x+1)(x+3)+2/(x+3)(x+5)+…+2/(x+99)(x+101)=?
2/(x+1)(x+3)+2/(x+3)(x+5)+…+2/(x+99)(x+101)=?
=2[1/(x+1)(x+3)+1/(x+3)(x+5)+…+1/(x+99)(x+101)]
=2{[2/(X+1)(X+3)+2/(X+3)(X+5)+2/(X+5)(X+7)+...+2/(X+99)(X+101)]/2}
=1/(X+1)-1/(X+3)+1/(X+3)-1/(X+5)+1/(X+5)-1/(X+7)+...+1/(X+99)-1/(X+101)
=1/(X+1)-1/(X+101)
=(X+101-X-1)/(X+1)(X+101)
=100/(X+1)(X+101)
=2{[2/(X+1)(X+3)+2/(X+3)(X+5)+2/(X+5)(X+7)+...+2/(X+99)(X+101)]/2}
=1/(X+1)-1/(X+3)+1/(X+3)-1/(X+5)+1/(X+5)-1/(X+7)+...+1/(X+99)-1/(X+101)
=1/(X+1)-1/(X+101)
=(X+101-X-1)/(X+1)(X+101)
=100/(X+1)(X+101)
分式:计算1/(x-1)+1/(x-1)(x-2)+1/(x-2)(x-3)+...+1/(x-99)(x-100)
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