$\displaystyle \begin{array}{{>{\displaystyle}l}} A=\sqrt{4+\sqrt{10+2\sqrt{5}}} +\sqrt{4-\sqrt{10+2\sqrt{5}}}\\ A^{2} =4+\sqrt{10+2\sqrt{5}} +4-\sqrt{10+2\sqrt{5}} +2\sqrt{\left( 4+\sqrt{10+2\sqrt{5}}\right)\left( 4-\sqrt{10+2\sqrt{5}}\right)}\\ A^{2} =8+2\sqrt{16-10+2\sqrt{5}}\\ A^{2} =8+2\sqrt{6+2\sqrt{5}}\\ A^{2} =8+2\sqrt{\left( 1+\sqrt{5}\right)^{2}}\\ A^{2} =8+2\left( 1+\sqrt{5}\right) =8+2+2\sqrt{5}\\ A^{2} =10+2\sqrt{5} \ \\ \rightarrow A=\sqrt{10+2\sqrt{5}}\\ B=\left(\sqrt{2} +1\right)\sqrt[3]{40\sqrt{2} -56} \ \\ B^{3} =\left(\sqrt{2} +1\right)^{3} .\left( 40\sqrt{2} -56\right)\\ B^{3} =\left( 7+5\sqrt{2}\right) .8\left( 5\sqrt{2} -7\right)\\ B^{3} =8( 50-49)\\ B^{3} =8\ \rightarrow B=2\ \\ C\sqrt{0,25\sqrt{961} +2\sqrt{10} +\sqrt{15} +\sqrt{6}}\\ =\sqrt{\frac{31}{4} +2\sqrt{10} +\sqrt{15} +\sqrt{6}}\\ =\sqrt{\frac{31+8\sqrt{10} +4\sqrt{15} +4\sqrt{6}}{4}}\\ =\sqrt{\frac{8+3+20+2.2\sqrt{2} .2\sqrt{5} +2.2\sqrt{3} .\sqrt{5} +2.2\sqrt{2}\sqrt{3}}{4}}\\ =\sqrt{\frac{\left( 2\sqrt{2} +2\sqrt{5} +\sqrt{3}\right)^{2}}{2^{2}}} =\frac{2\sqrt{2} +2\sqrt{5} +\sqrt{3}}{2} \ \\ \\ \\ \\ \end{array}$
Còn câu D mình chịu
