Question

[_Banhdayyy_]

Rút gọn biểu thức (làm chi tiết)

Answer 1

[Nguyễn Hoàng Minh]

\(e,\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}\\ =\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2\\ =\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}-\sqrt{2}\right)^3\)

\(f,\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\\ =36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}=36-36\sqrt{2}+27\sqrt{3}\)

\(g,\left(2+\sqrt{3}-\sqrt{2}\right)\left(2-\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{2}\right)\sqrt{3-2\sqrt{2}}\\ =\left[\left(2-\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2\right]\left(3+\sqrt{2}\right)\sqrt{\left(\sqrt{2}-1\right)^2}\\ =\left(3-4\sqrt{2}\right)\left(3+\sqrt{2}\right)\left(\sqrt{2}-1\right)\\ =\left(1-9\sqrt{2}\right)\left(\sqrt{2}-1\right)\\ =10\sqrt{2}-37\)

\(h,A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\\ A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ A^2=8+2\sqrt{6-2\sqrt{5}}\\ A^2=8+2\left(\sqrt{5}-1\right)\\ A^2=6+2\sqrt{5}\\ A=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)