a,
$ĐKXĐ:x>1$
$M=(\dfrac{1}{\sqrt[]x-1}-\dfrac{2.\sqrt[]x}{\sqrt[]x(x+1)-(x+1)}):\dfrac{\sqrt[]x+x+1}{x+1}$
$=(\dfrac{x+1}{(x+1)(\sqrt[]x-1)}-\dfrac{2.\sqrt[]x}{(x+1)(\sqrt[]x-1)}).\dfrac{x+1}{\sqrt[]x+x+1}$ $=\dfrac{x-2.\sqrt[]x+1}{\sqrt[]x-1}.\dfrac{1}{\sqrt[]x+x+1}$
$=\dfrac{(\sqrt[]x-1)^2}{\sqrt[]x-1}.\dfrac{1}{\sqrt[]x+x+1}$ $=(\sqrt[]x-1).\dfrac{1}{\sqrt[]x+x+1}$ $=\dfrac{\sqrt[]x-1}{\sqrt[]x+x+1}$
b, Theo câu a ta có: $M=\dfrac{\sqrt[]x-1}{\sqrt[]x+x+1}$
$⇒M>0⇔\dfrac{\sqrt[]x-1}{\sqrt[]x+x+1}>0$
Mà $x+\sqrt[]x+1=(\sqrt[]x^2)+2.\dfrac{1}{2}.\sqrt[]x+\dfrac{1}{4}+\dfrac{3}{4}=(\sqrt[]x+\dfrac{1}{2})^2+\dfrac{3}{4}>0∀x$
$⇒\sqrt[]x-1>0$
$⇔\sqrt[]x>1$
$⇔x>1$
Vậy $x>1$ thì $M>0$
c,
Ta có: $x=9-4.\sqrt[]5$
$⇒\sqrt[]x=\sqrt[]{9-4.\sqrt[]5}$
$⇒\sqrt[]x=\sqrt[]{5-2.2.\sqrt[]5+4}$
$⇒\sqrt[]x=\sqrt[]{(\sqrt[]5-2)^2}$
$⇒\sqrt[]x=\sqrt[]5-2$
Khi đó $M=\dfrac{\sqrt[]x-1}{\sqrt[]x+x+1}$
$=\dfrac{\sqrt[]5-2-1}{\sqrt[]5-2+9-4.\sqrt[]5+1}$
$=\dfrac{(\sqrt[]5-3)}{8-3.\sqrt[]5}$
