Question

RÚT GỌN

a) \(\dfrac{x+3}{x-9}\) + \(\dfrac{2}{\sqrt{x}+3}\) \(-\dfrac{1}{\sqrt{x}-3}\)\(\left(x\ge0\right),x\ne9\))

b) \(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\)\(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)(\(x\ge0\))

CÁC BÁC GIÚP EM VỚI, EM SẮP KTRA RỒI HUHUU

Answer 1

a/ \(\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}=\dfrac{x+3+2\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)}{x-9}=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{x-9}=\dfrac{x-\sqrt{x}-6}{x-9}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)}{x-9}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

b/ \(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{3\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\)

P/s: câu b đề sai phải không bạn, mk nghĩ ngoài dấu ngoặc là phép chia thì đúng hơn