Mng giải giúp mk nha . Mk đnag cần gấp . Càng nhanh càng tốt nha . Tks mng.

Bài 2: 1) $A=\frac1{\sqrt2-\sqrt7}-\frac1{\sqrt2+\sqrt7}$ Điều kiện xác định: $\sqrt{2} \neq \sqrt{7}$. Ta có: \[ A = \frac{1}{\sqrt{2} - \sqrt{7}} - \frac{1}{\sqrt{2} + \sqrt{7}} \] Nhân liên hợp: \[ A = \frac{\sqrt{2} + \sqrt{7}}{(\sqrt{2} - \sqrt{7})(\sqrt{2} + \sqrt{7})} - \frac{\sqrt{2} - \sqrt{7}}{(\sqrt{2} + \sqrt{7})(\sqrt{2} - \sqrt{7})} \] \[ A = \frac{\sqrt{2} + \sqrt{7}}{2 - 7} - \frac{\sqrt{2} - \sqrt{7}}{2 - 7} \] \[ A = \frac{\sqrt{2} + \sqrt{7} - (\sqrt{2} - \sqrt{7})}{-5} \] \[ A = \frac{\sqrt{2} + \sqrt{7} - \sqrt{2} + \sqrt{7}}{-5} \] \[ A = \frac{2\sqrt{7}}{-5} \] \[ A = -\frac{2\sqrt{7}}{5} \] 2) $A=\frac{5-2\sqrt5}{\sqrt5-2}-\sqrt{6-2\sqrt5}$ Điều kiện xác định: $\sqrt{5} \neq 2$. Ta có: \[ A = \frac{5 - 2\sqrt{5}}{\sqrt{5} - 2} - \sqrt{6 - 2\sqrt{5}} \] Nhân liên hợp: \[ A = \frac{(5 - 2\sqrt{5})(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} - \sqrt{6 - 2\sqrt{5}} \] \[ A = \frac{5\sqrt{5} + 10 - 10 - 4\sqrt{5}}{5 - 4} - \sqrt{6 - 2\sqrt{5}} \] \[ A = \frac{\sqrt{5}}{1} - \sqrt{6 - 2\sqrt{5}} \] \[ A = \sqrt{5} - \sqrt{6 - 2\sqrt{5}} \] \[ A = \sqrt{5} - \sqrt{(\sqrt{5} - 1)^2} \] \[ A = \sqrt{5} - (\sqrt{5} - 1) \] \[ A = \sqrt{5} - \sqrt{5} + 1 \] \[ A = 1 \] 3) $A=\sqrt5-\frac8{\sqrt5+1}+\frac{2\sqrt5-5}{2-\sqrt5}$ Điều kiện xác định: $\sqrt{5} \neq -1$, $\sqrt{5} \neq 2$. Ta có: \[ A = \sqrt{5} - \frac{8}{\sqrt{5} + 1} + \frac{2\sqrt{5} - 5}{2 - \sqrt{5}} \] Nhân liên hợp: \[ A = \sqrt{5} - \frac{8(\sqrt{5} - 1)}{(\sqrt{5} + 1)(\sqrt{5} - 1)} + \frac{(2\sqrt{5} - 5)(2 + \sqrt{5})}{(2 - \sqrt{5})(2 + \sqrt{5})} \] \[ A = \sqrt{5} - \frac{8(\sqrt{5} - 1)}{5 - 1} + \frac{(2\sqrt{5} - 5)(2 + \sqrt{5})}{4 - 5} \] \[ A = \sqrt{5} - \frac{8(\sqrt{5} - 1)}{4} + \frac{(2\sqrt{5} - 5)(2 + \sqrt{5})}{-1} \] \[ A = \sqrt{5} - 2(\sqrt{5} - 1) - (2\sqrt{5} - 5)(2 + \sqrt{5}) \] \[ A = \sqrt{5} - 2\sqrt{5} + 2 - (4\sqrt{5} + 10 - 10 - 5\sqrt{5}) \] \[ A = \sqrt{5} - 2\sqrt{5} + 2 - (-\sqrt{5}) \] \[ A = \sqrt{5} - 2\sqrt{5} + 2 + \sqrt{5} \] \[ A = 2 \] 4) $A=\frac{\sqrt6-\sqrt3}{\sqrt2-1}-\sqrt{(1-\sqrt3)^2}$ Điều kiện xác định: $\sqrt{2} \neq 1$. Ta có: \[ A = \frac{\sqrt{6} - \sqrt{3}}{\sqrt{2} - 1} - \sqrt{(1 - \sqrt{3})^2} \] Nhân liên hợp: \[ A = \frac{(\sqrt{6} - \sqrt{3})(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} - |1 - \sqrt{3}| \] \[ A = \frac{\sqrt{12} + \sqrt{6} - \sqrt{6} - \sqrt{3}}{2 - 1} - |\sqrt{3} - 1| \] \[ A = \frac{2\sqrt{3} - \sqrt{3}}{1} - (\sqrt{3} - 1) \] \[ A = \sqrt{3} - \sqrt{3} + 1 \] \[ A = 1 \] 5) $A=\sqrt{(1-\sqrt3)^2}+\sqrt{12}-\frac6{\sqrt3-1}$ Điều kiện xác định: $\sqrt{3} \neq 1$. Ta có: \[ A = \sqrt{(1 - \sqrt{3})^2} + \sqrt{12} - \frac{6}{\sqrt{3} - 1} \] \[ A = |1 - \sqrt{3}| + 2\sqrt{3} - \frac{6}{\sqrt{3} - 1} \] \[ A = \sqrt{3} - 1 + 2\sqrt{3} - \frac{6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \] \[ A = \sqrt{3} - 1 + 2\sqrt{3} - \frac{6(\sqrt{3} + 1)}{3 - 1} \] \[ A = \sqrt{3} - 1 + 2\sqrt{3} - \frac{6(\sqrt{3} + 1)}{2} \] \[ A = \sqrt{3} - 1 + 2\sqrt{3} - 3(\sqrt{3} + 1) \] \[ A = \sqrt{3} - 1 + 2\sqrt{3} - 3\sqrt{3} - 3 \] \[ A = \sqrt{3} - 1 + 2\sqrt{3} - 3\sqrt{3} - 3 \] \[ A = -4 \] 6) $A=\frac{\sqrt{15}-\sqrt3}{\sqrt5-1}+\sqrt{(1-\sqrt3)^2}-2\sqrt3$ Điều kiện xác định: $\sqrt{5} \neq 1$. Ta có: \[ A = \frac{\sqrt{15} - \sqrt{3}}{\sqrt{5} - 1} + \sqrt{(1 - \sqrt{3})^2} - 2\sqrt{3} \] Nhân liên hợp: \[ A = \frac{(\sqrt{15} - \sqrt{3})(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} + |1 - \sqrt{3}| - 2\sqrt{3} \] \[ A = \frac{\sqrt{75} + \sqrt{15} - \sqrt{15} - \sqrt{3}}{5 - 1} + \sqrt{3} - 1 - 2\sqrt{3} \] \[ A = \frac{5\sqrt{3} - \sqrt{3}}{4} + \sqrt{3} - 1 - 2\sqrt{3} \] \[ A = \frac{4\sqrt{3}}{4} + \sqrt{3} - 1 - 2\sqrt{3} \] \[ A = \sqrt{3} + \sqrt{3} - 1 - 2\sqrt{3} \] \[ A = -1 \] 7) $A=\sqrt{48}-\frac{\sqrt{21}-\sqrt{15}}{\sqrt7-\sqrt5}+\frac2{\sqrt3+1}$ Điều kiện xác định: $\sqrt{7} \neq \sqrt{5}$, $\sqrt{3} \neq -1$. Ta có: \[ A = \sqrt{48} - \frac{\sqrt{21} - \sqrt{15}}{\sqrt{7} - \sqrt{5}} + \frac{2}{\sqrt{3} + 1} \] Nhân liên hợp: \[ A = 4\sqrt{3} - \frac{(\sqrt{21} - \sqrt{15})(\sqrt{7} + \sqrt{5})}{(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5})} + \frac{2(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \] \[ A = 4\sqrt{3} - \frac{\sqrt{147} + \sqrt{105} - \sqrt{105} - \sqrt{75}}{7 - 5} + \frac{2(\sqrt{3} - 1)}{3 - 1} \] \[ A = 4\sqrt{3} - \frac{7\sqrt{3} - 5\sqrt{3}}{2} + \frac{2(\sqrt{3} - 1)}{2} \] \[ A = 4\sqrt{3} - \frac{2\sqrt{3}}{2} + \sqrt{3} - 1 \] \[ A = 4\sqrt{3} - \sqrt{3} + \sqrt{3} - 1 \] \[ A = 4\sqrt{3} - 1 \] 8) $A=\frac{6+2\sqrt3}{\sqrt3+1}-\frac2{\sqrt5-\sqrt3}+\sqrt{8-2\sqrt{15}}$ Điều kiện xác định: $\sqrt{3} \neq -1$, $\sqrt{5} \neq \sqrt{3}$. Ta có: \[ A = \frac{6 + 2\sqrt{3}}{\sqrt{3} + 1} - \frac{2}{\sqrt{5} - \sqrt{3}} + \sqrt{8 - 2\sqrt{15}} \] Nhân liên hợp: \[ A = \frac{(6 + 2\sqrt{3})(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} - \frac{2(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} + \sqrt{(\sqrt{5} - \sqrt{3})^2} \] \[ A = \frac{6\sqrt{3} - 6 + 6 - 2\sqrt{3}}{3 - 1} - \frac{2(\sqrt{5} + \sqrt{3})}{5 - 3} + \sqrt{5} - \sqrt{3} \] \[ A = \frac{4\sqrt{3}}{2} - \frac{2(\sqrt{5} + \sqrt{3})}{2} + \sqrt{5} - \sqrt{3} \] \[ A = 2\sqrt{3} - (\sqrt{5} + \sqrt{3}) + \sqrt{5} - \sqrt{3} \] \[ A = 2\sqrt{3} - \sqrt{5} - \sqrt{3} + \sqrt{5} - \sqrt{3} \] \[ A = 0 \] 9) $A=\frac4{\sqrt3+1}+\frac1{\sqrt3-2}+\frac6{\sqrt3-3}$ Điều kiện xác định: $\sqrt{3} \neq -1$, $\sqrt{3} \neq 2$, $\sqrt{3} \neq 3$. Ta có: \[ A = \frac{4}{\sqrt{3} + 1} + \frac{1}{\sqrt{3} - 2} + \frac{6}{\sqrt{3} - 3} \] Nhân liên hợp: \[ A = \frac{4(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} + \frac{1(\sqrt{3} + 2)}{(\sqrt{3} - 2)(\sqrt{3} + 2)} + \frac{6(\sqrt{3} + 3)}{(\sqrt{3} - 3)(\sqrt{3} + 3)} \] \[ A = \frac{4(\sqrt{3} - 1)}{3 - 1} + \frac{1(\sqrt{3} + 2)}{3 - 4} + \frac{6(\sqrt{3} + 3)}{3 - 9} \] \[ A = \frac{4(\sqrt{3} - 1)}{2} + \frac{\sqrt{3} + 2}{-1} + \frac{6(\sqrt{3} + 3)}{-6} \] \[ A = 2(\sqrt{3} - 1) - (\sqrt{3} + 2) - (\sqrt{3} + 3) \] \[ A = 2\sqrt{3} - 2 - \sqrt{3} - 2 - \sqrt{3} - 3 \] \[ A = -7 \] 10) $A=(\frac{15}{\sqrt6+1}+\frac2{\sqrt6-2}-\frac6{3-\sqrt6})(2\sqrt6+7)$ Điều kiện xác định: $\sqrt{6} \neq -1$, $\sqrt{6} \neq 2$, $\sqrt{6} \neq 3$. Ta có: \[ A = \left( \frac{15}{\sqrt{6} + 1} + \frac{2}{\sqrt{6} - 2} - \frac{6}{3 - \sqrt{6}} \right) (2\sqrt{6} + 7) \] Nhân liên hợp: \[ A = \left( \frac{15(\sqrt{6} - 1)}{(\sqrt{6} + 1)(\sqrt{6} - 1)} + \frac{2(\sqrt{6} + 2)}{(\sqrt{6} - 2)(\sqrt{6} + 2)} - \frac{6(3 + \sqrt{6})}{(3 - \sqrt{6})(3 + \sqrt{6})} \right) (2\sqrt{6} + 7) \] \[ A = \left( \frac{15(\sqrt{6} - 1)}{6 - 1} + \frac{2(\sqrt{6} + 2)}{6 - 4} - \frac{6(3 + \sqrt{6})}{9 - 6} \right) (2\sqrt{6} + 7) \] \[ A = \left( \frac{15(\sqrt{6} - 1)}{5} + \frac{2(\sqrt{6} + 2)}{2} - \frac{6(3 + \sqrt{6})}{3} \right) (2\sqrt{6} + 7) \] \[ A = \left( 3(\sqrt{6} - 1) + (\sqrt{6} + 2) - 2(3 + \sqrt{6}) \right) (2\sqrt{6} + 7) \] \[ A = \left( 3\sqrt{6} - 3 + \sqrt{6} + 2 - 6 - 2\sqrt{6} \right) (2\sqrt{6} + 7) \] \[ A = \left( 2\sqrt{6} - 7 \right) (2\sqrt{6} + 7) \] \[ A = (2\sqrt{6})^2 - 7^2 \] \[ A = 24 - 49 \] \[ A = -25 \] 11) $A=\frac{2+\sqrt3}{2-\sqrt3}+\frac{2-\sqrt3}{2+\sqrt3}$ Điều kiện xác định: $2 \neq \sqrt{3}$. Ta có: \[ A = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \] Nhân liên hợp: \[ A = \frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} + \frac{(2 - \sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \] \[ A = \frac{(2 + \sqrt{3})^2}{4 - 3} + \frac{(2 - \sqrt{3})^2}{4 - 3} \] \[ A = (2 + \sqrt{3})^2 + (2 - \sqrt{3})^2 \] \[ A = 4 + 4\sqrt{3} + 3 + 4 - 4\sqrt{3} + 3 \] \[ A = 14 \] 12) $A=\sqrt{\frac{3+\sqrt5}{3-\sqrt5}}+\sqrt{\frac{3-\sqrt5}{3+\sqrt5}}$ Điều kiện xác định: $3 \neq \sqrt{5}$. Ta có: \[ A = \sqrt{\frac{3 + \sqrt{5}}{3 - \sqrt{5}}} + \sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}} \] Nhân liên hợp: \[ A = \sqrt{\frac{(3 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}} + \sqrt{\frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}} \] \[ A = \sqrt{\frac{(3 + \sqrt{5})^2}{9 - 5}} + \sqrt{\frac{(3 - \sqrt{5})^2}{9 - 5}} \] \[ A = \sqrt{\frac{(3 + \sqrt{5})^2}{4}} + \sqrt{\frac{(3 - \sqrt{5})^2}{4}} \] \[ A = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} \] \[ A = \frac{3 + \sqrt{5} + 3 - \sqrt{5}}{2} \] \[ A = \frac{6}{2} \] \[ A = 3 \] Đáp số: 1) $A = -\frac{2\sqrt{7}}{5}$ 2) $A = 1$ 3) $A = 2$ 4) $A = 1$ 5) $A = -4$ 6) $A = -1$ 7) $A = 4\sqrt{3} - 1$ 8) $A = 0$ 9) $A = -7$ 10) $A = -25$ 11) $A = 14$ 12) $A = 3$