Question

[Hương Phùng]

B1: rút gọn:

a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)

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

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

d, \(\dfrac{x^2-5}{x+\sqrt{5}}\)         x khác \(-\sqrt{5}\)

e, \(\dfrac{x^2+2\sqrt{2}x+2}{x+\sqrt{2}}\)       x khác \(-\sqrt{2}\)

g, \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

giúp em với ạ , em cảm ơn 

Answer 1

[Kiêm Hùng]

Answer 2

[An Thy]

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

\(=x-4+x-4\left(x>4\right)=2x-8\)

d) \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)

e) \(\dfrac{x^2+2\sqrt{2}x+2}{x+\sqrt{2}}=\dfrac{\left(x+\sqrt{2}\right)^2}{x+\sqrt{2}}=x+\sqrt{2}\)

g) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{1}{\sqrt{2}}\)