$A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}$
a, rút gọn A
b, Tính giá trị A khi x=$6+2\sqrt{5}$
c, Tìm x để A=$\frac{1}{2}$
d, Tìm x để A>1
e, Tìm $x\in Z$ để $A\in Z$
(làm chi tiết)
Mấy bài này thực chất không khó
A=$\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(x-5\sqrt{x}+6)}+\frac{(2\sqrt{x}+1)(\sqrt{x}-2)}{x-5\sqrt{x}+6}=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{x-5\sqrt{x}+6}=\frac{x-\sqrt{x}-2}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}(x\geq 0,x\neq 4,x\neq 9)$
b.$A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{5}+2}{\sqrt{5}-2}=\frac{(\sqrt{5}+2)^2}{1}=(\sqrt{5}+2)^2$
c.$\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{1}{2}<=>\sqrt{x}=-5=>x\epsilon \phi$
d.$\frac{\sqrt{x}+1}{\sqrt{x}-3}>1<=>\frac{4}{\sqrt{x}-3}>0=>\sqrt{x}>3=>x>9$
e.$A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}=>(\sqrt{x}-3)\epsilon {\pm 1,\pm 2,\pm 4}=>\sqrt{x}=2,4,1,5,7=>x=4,16,1,25,49$