Question

Rút gọn các biểu thức:

a) \(\sqrt{4-2\sqrt{3}}\)-\(\sqrt{3}\)

b) \(\sqrt{11+6\sqrt{2}}\)-3+\(\sqrt{2}\)

c) \(\sqrt{9x^2}\)-2x với x<0

d) x-4+\(\sqrt{16-8x+x^2x^2}\) với x>4

Answer 1

a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)

b,\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}=\sqrt{2}+3-3+\sqrt{2}=2\sqrt{2}\)

c, \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x=3x-2x=x\)

d, câu này sai đề rồi , nếu sửa lại phải như này :

\(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}=x-4+4-x=0\)

Answer 2

a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)

b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)

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

c) \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x\) = \(\left|3x\right|-2x=-3x-2x\) (x < 0)

= \(-5x\)

d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)

= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) ( \(x>4\))

= \(2x-8\)