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(1/x+1%1/1%x) 怎么解?

是不是(1/(x+1)-1/(1-x))? 1/(x+1)-1/(1-x) =1/(x+1)+1/(x-1) =(x-1)/((x+1)(x-1)) + (x+1)/((x+1)(x-1))\ =((x-1)+ (x+1))/((x+1)(x-1)) =2x/((x+1)(x-1)) =2x/(x^2-1)

1/4(x+1)=1/3(x-1) 同乘以12,得 3(x+1)=4(x-1) 3x+3=4x-4 4x-3x=3+4 x=7

=2000x[(P/A.1%,11)+1] =2000x(10.368+1) =22736

(1-x)/(1+x)>=1 (1-x)/(1+x)-1>=0 通分 [(1-x)-1-x]/(1+x)>=0 -2x/(1+x)>=0 x/(1+x)

两边乘6x(x+5)得 6(x+5)+6x=x(x+5) 整理得 x^2-7x-30=0 (x-10)(x+3)=0 x=10,x=-3 经检验两者都是原方程的根

解:(1+x)/(1-x)>1 当1-x>0时,即x1-x 2x>0 x>0 综合一下,0

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