M=(($\frac{1}{a-√a}$ +$\frac{1}{√a -1}$): $\frac{√a+1}{a-2√a+1}$
=($\frac{1}{√a(√a-1)}$ + $\frac{1}{√a-1}$).$\frac{(√a-1)^2}{√a+1}$
=($\frac{1}{√a(√a-1)}$ + $\frac{√a}{√a(√a-1)}$).$\frac{(√a-1)^2}{√a+1}$
= $\frac{1+√a}{√a(√a-1)}$+.$\frac{(√a-1)^2}{√a+1}$
= $\frac{√a−1}{√a}$
Vậy M=$\frac{√a−1}{√a}$
Để M>0 thì $\frac{√a−1}{√a}$>0
=>√a−1>0
=> a>1 ( vì a>0)
Vậy a>±1 để M>0
Để M<1 thì $\frac{√a−1}{√a}$ <1
=> $\frac{√a−1}{√a}$ -1 <0
=> $\frac{√a−1}{√a}$ - $\frac{√a}{√a}$ <0
=> $\frac{-1}{√a}$ <0
=> √a>0
=> a>0
Vậy a>0 để M<1
