TOÁN ĐẠI 9: Tính giá tri biểu thức ( khó)

N

nguyenbahiep1

Gới ý cách làm như sau


[laTEX]\sqrt{2+\sqrt{3}} =\sqrt{\frac{4+2\sqrt{3}}{2}} = \frac{1+\sqrt{3}}{\sqrt{2}} \\ \\ \Rightarrow \sqrt{2+\sqrt{3}} = \frac{\sqrt{6}+\sqrt{2}}{2} \\ \\ x = \sqrt{2+\frac{\sqrt{6}+\sqrt{2}}{2}} - \sqrt{3(2-\frac{\sqrt{6}+\sqrt{2}}{2})} \\ \\ x^2 = 8 -(\sqrt{6}+\sqrt{2}) - 2\sqrt{3}.\sqrt{(2+\frac{\sqrt{6}+\sqrt{2}}{2})(2+\frac{\sqrt{6}-\sqrt{2}}{2})} \\ \\ x^2 = 8 - (\sqrt{6}+\sqrt{2}) -2\sqrt{3}.\sqrt{2-\sqrt{3}} \\ \\ x^2 = 8 - (\sqrt{6}+\sqrt{2}) -2.\sqrt{3}\frac{\sqrt{6}-\sqrt{2}}{2} \\ \\ x^2 = 8 - (\sqrt{6}+\sqrt{2}) + \sqrt{6} - 3\sqrt{2} = 8 - 4\sqrt{2}[/laTEX]
 
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