b/ $P\,=\dfrac{-5\sqrt x+2}{\sqrt x+3}\\\quad =\dfrac{-5\sqrt x-15+17}{\sqrt x+3}\\\quad =\dfrac{-5(\sqrt x+3)+17}{\sqrt x+3}\\\quad =\dfrac{-5(\sqrt x+3)}{\sqrt x+3}+\dfrac{17}{\sqrt x+3}\\\quad =-5+\dfrac{17}{\sqrt x+3}$
Ta có: $\sqrt x\ge 0$
$↔\sqrt x+3\ge 3\\↔\dfrac{17}{\sqrt x+3}\le \dfrac{17}{3}\\↔-5+\dfrac{17}{\sqrt x+3}\le \dfrac{2}{3}$
mà $\dfrac{2}{3}<1$
$→-5+\dfrac{17}{\sqrt x+3}<1\\↔P<1$
Vậy $P<1$
c/ $x=4-2\sqrt 3\\\quad =3-2\sqrt 3+1\\\quad =(\sqrt 3-1)^2$
Với $x=4-2\sqrt 3$ (tmđk) thì
$P\,=\dfrac{-5\sqrt{(\sqrt 3-1)^2}+2}{\sqrt{(\sqrt 3-1)^2}}\\\quad =\dfrac{-5|\sqrt 3-1|+2}{|\sqrt 3-1|}\\\quad =\dfrac{-5(\sqrt 3-1)+2}{\sqrt 3-1}\\\quad =\dfrac{-5\sqrt 3+5+2}{\sqrt 3-1}\\\quad =\dfrac{7-5\sqrt 3}{\sqrt 3-1}\\\quad =\dfrac{(7-5\sqrt 3)(\sqrt 3+1)}{(\sqrt 3-1)(\sqrt 3+1)}\\\quad =\dfrac{7-5\sqrt 3.\sqrt 3-5\sqrt 3+7\sqrt 3}{(\sqrt 3)^2-1^2}\\\quad =\dfrac{7-15+2\sqrt 3}{3-1}\\\quad =\dfrac{-8+2\sqrt 3}{2}$
Vậy $P=\dfrac{-8+2\sqrt 3}{2}$ với $x=4-2\sqrt 3$
d/ $P>0\\↔\dfrac{-5\sqrt x+2}{\sqrt x+3}>0$
mà $\sqrt x+3\ge 0>0$
$→-5\sqrt x+2>0\\↔-5\sqrt x>-2\\↔5\sqrt x<2\\↔\sqrt x<\dfrac{2}{5}\\↔x<\dfrac{4}{25}$
Kết hợp điều kiện: $x\ge 0,x\ne 1$
$→0\le x<\dfrac{4}{25}$
Vậy $P>0$ thì $0\le x<\dfrac{4}{25}$
