Bài 2:
$M=\dfrac{\sqrt{x}-3}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{10-6\sqrt{x}}{x-2\sqrt{x}} (x > 0 ; x \neq 4)$
$=\dfrac{(\sqrt{x}-3)(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}-\dfrac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-2)}-\dfrac{10-6\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}$
$=\dfrac{x-2\sqrt{x}-3\sqrt{x}+6-x+\sqrt{x}-10+6\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}$
$=\dfrac{2\sqrt{x}-4}{\sqrt{x}(\sqrt{x}-2)}$
$=\dfrac{2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}$
$=\dfrac{2}{\sqrt{x}}$
Vậy $M=\dfrac{2}{\sqrt{x}}$ với $x > 0 ; x \neq 4$
Bài 3 :
Ta có : $y=\dfrac{4}{5}x+20$
$⇔48=\dfrac{4}{5}x+20$
$⇔\dfrac{4}{5}x=28$
$⇔x=35$
Vậy khối lượng hành lý quá cước là $35 kg$
