bài 1: Tính:
\(\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)+\(\dfrac{3+6\sqrt{3}}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}\)
Baì 2 : rút gọn
cho A=\((\) \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\) \()\): \((\dfrac{4\sqrt{x}-8}{1-x})\)
a. tìm x để A xác định
b. Rút gọn A
a/ ĐKXĐ: x>= 0 ; x khác 1
b/ \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\dfrac{4\sqrt{x}-8}{1-x}\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}-\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}-\dfrac{8\sqrt{x}}{x-1}\right):\dfrac{8-4\sqrt{x}}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{2}-1-8\sqrt{x}}{x-1}\cdot\dfrac{x-1}{8-4\sqrt{x}}\)
\(=\dfrac{-4\sqrt{x}}{x-1}\cdot\dfrac{x-1}{4\left(2-\sqrt{x}\right)}=\dfrac{-4\sqrt{x}}{4\left(2-\sqrt{x}\right)}=-\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
Làm nốt bài 1 ::v
\(\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\dfrac{3+6\sqrt{3}}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}=\dfrac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}=6-\dfrac{13}{\sqrt{3}+4}=\dfrac{11+6\sqrt{3}}{\sqrt{3}+4}\)
