Sử dụng các phép biến đổi đơn giản tính giá trị biểu thức

4. Rút gọn các biểu thức sau:

a, M = $\sqrt{125}-3\sqrt{45}+2\sqrt{20}$

b, N = $3\sqrt{\frac{1}{2}}+\sqrt{4,5}-\sqrt{12,5}$

c, P = $\sqrt{18}-\sqrt{8}+\sqrt{50}-\sqrt{578}+\sqrt{128}-\sqrt{242}+\sqrt{72}$

d, Q = $\sqrt{\frac{3}{2}-\frac{\sqrt{5}}{2}}+\sqrt{\frac{49}{9}+\frac{4}{3}\sqrt{5}}-\sqrt{\frac{3}{2}+\frac{\sqrt{5}}{2}}$

e, E = $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{3}{\sqrt{5}-\sqrt{2}}$

f, F = $\left ( \frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}} \right ):\frac{1}{\sqrt{5}-\sqrt{2}}$


4. a, M = $\sqrt{125}-3\sqrt{45}+2\sqrt{20}$ =  $\sqrt{25.5}-3\sqrt{9.5}+2\sqrt{4.5}$

           = $5\sqrt{5}-3.3\sqrt{5}+2.2\sqrt{5}$ = 0

b, N = $3\sqrt{\frac{1}{2}}+\sqrt{4,5}-\sqrt{12,5}$ = $3\sqrt{\frac{1}{2}}+\sqrt{\frac{9}{2}}-\sqrt{\frac{25}{2}}$ 

       = $3\sqrt{\frac{2}{2.2}}+\sqrt{\frac{9.2}{2.2}}-\sqrt{\frac{25.2}{2.2}}$ =  $\frac{3\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}-\frac{5\sqrt{2}}{2}$

       = $\frac{\sqrt{2}}{2}$

c, P = $\sqrt{18}-\sqrt{8}+\sqrt{50}-\sqrt{578}+\sqrt{128}-\sqrt{242}+\sqrt{72}$

       = $\sqrt{9.2}-\sqrt{4.2}+\sqrt{25.2}-\sqrt{289.2}+\sqrt{64.2}-\sqrt{121.2}+\sqrt{36.2}$

       = $3\sqrt{2}-4\sqrt{2}+5\sqrt{2}-17\sqrt{2}+8\sqrt{2}-11\sqrt{2}+6\sqrt{2}$

       = $-8\sqrt{2}$

d, Q = $\sqrt{\frac{3}{2}-\frac{\sqrt{5}}{2}}+\sqrt{\frac{49}{9}+\frac{4}{3}\sqrt{5}}-\sqrt{\frac{3}{2}+\frac{\sqrt{5}}{2}}$

       = $\sqrt{\frac{3-\sqrt{5}}{2}}+\sqrt{\frac{49+12\sqrt{5}}{9}}-\sqrt{\frac{3+\sqrt{5}}{2}}$

       = $\sqrt{\frac{6-2\sqrt{5}}{2.2}}+\sqrt{\frac{49+12\sqrt{5}}{9}}-\sqrt{\frac{6+2\sqrt{5}}{2.2}}$

       = $\sqrt{\frac{(\sqrt{5}-1)^{2}}{2.2}}+\sqrt{\frac{(3\sqrt{5}+2)^{2}}{9}}-\sqrt{\frac{(\sqrt{5}+1)^{2}}{2.2}}$

       = $\frac{\sqrt{5}-1}{2}+\frac{3\sqrt{5}+2}{3}-\frac{\sqrt{5}+1}{2}$

       = $\frac{3\sqrt{5}+2}{3}$

e, E = $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{3}{\sqrt{5}-\sqrt{2}}$ = $\frac{3.(\sqrt{5}-\sqrt{2})+3.(\sqrt{5}+\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$ 

=  $\frac{3.(\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{5-2}$ = $\frac{3.2\sqrt{5}}{3}$ = $2\sqrt{5}$

f, F = $\left ( \frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}} \right ):\frac{1}{\sqrt{5}-\sqrt{2}}$

      = $\left ( \frac{(\sqrt{6}-\sqrt{2})(1+\sqrt{3})}{1-3}-\frac{\sqrt{5}.\sqrt{5}}{\sqrt{5}} \right ):\frac{1}{\sqrt{5}-\sqrt{2}}$

      = $\left ( \frac{\sqrt{6}+\sqrt{18}-\sqrt{2}-\sqrt{6}}{-2}-\sqrt{5} \right ):\frac{1}{\sqrt{5}-\sqrt{2}}$

      = $\left ( \frac{3\sqrt{2}-\sqrt{2}}{-2}-\sqrt{5} \right ):\frac{1}{\sqrt{5}-\sqrt{2}}$

      = $\left (-\sqrt{2}-\sqrt{5} \right ):\frac{1}{\sqrt{5}-\sqrt{2}}$

      = $-(\sqrt{2}+\sqrt{5}).(\sqrt{5}-\sqrt{2})$ = -(5 - 2 ) = -3


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