Question

[Nguyễn Hữu Minh Đức]

Em cần chạy deadline mong mn giúp nhanh ạ

Answer 1

[Tử Nguyệt Hàn]

1)
\(=\left(6\sqrt{2}-5\sqrt{2}+3\sqrt{2}\right):\sqrt{2}=4\sqrt{2}:\sqrt{2}=4\)
2)
\(=16a-2a+5a=19a\)
3)
\(=\dfrac{2\left(2-\sqrt{3}\right)}{\left(2 +\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\dfrac{3\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(\sqrt{3}+2\right)}\)
\(=4-2\sqrt{3}-6-3\sqrt{3}=-2-5\sqrt{3}\)

 

Answer 2

[Nguyễn Hoàng Minh]

\(1,=\left(6\sqrt{2}-5\sqrt{2}+3\sqrt{2}\right):\sqrt{2}=4\sqrt{2}:\sqrt{2}=4\\ 2,=\left|4a\right|-2\left|a\right|+5a=4a-2a+5a=7a\left(a\ge0\right)\\ 3,=\dfrac{4-2\sqrt{3}-6-3\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\dfrac{-5\sqrt{3}-2}{4-3}=-5\sqrt{3}-2\\ 4,=\sqrt{\left(3-\sqrt{5}\right)^2\left(3+\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)^2\left(3-\sqrt{5}\right)}\\ =\sqrt{4\left(3-\sqrt{5}\right)}+\sqrt{4\left(3+\sqrt{5}\right)}\\ =2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=2\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\\ =2\left(\sqrt{\dfrac{6-2\sqrt{5}}{2}}+\sqrt{\dfrac{6+2\sqrt{5}}{2}}\right)\\ =2\left(\dfrac{\sqrt{5}-1}{\sqrt{2}}+\dfrac{\sqrt{5}+1}{\sqrt{2}}\right)\\ =2\cdot\dfrac{2\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)