Question

B1: Cho b.thức B = \(\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\)
a) Tìm đk để B có nghĩa
b)Rút gọn B
B3: Tìm Min, Max của b.thức sau
A= \(\sqrt{x-2004}+\sqrt{2005-x}\)
B4: Rút gọn
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3 }} \)
b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
c)\(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)

Answer 1

Bài 4:

a: \(=2-\sqrt{3}+\sqrt{3}-1=1\)

b: \(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)

\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

c: \(=\dfrac{\left(15\cdot10\sqrt{2}-3\cdot15\sqrt{2}+2\cdot5\sqrt{2}\right)}{\sqrt{10}}\)

\(=15\cdot\sqrt{20}-3\cdot\sqrt{45}+2\cdot\sqrt{5}\)

\(=30\sqrt{5}-9\sqrt{5}+2\sqrt{5}=33\sqrt{5}\)