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 
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) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{1}{\sqrt{2}}\)
a) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
\(=\sqrt{3}-1-\sqrt{3}\)
=-1
b) Ta có: \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
c) Ta có: \(x-4+\sqrt{x^2-8x+16}\)
\(=x-4+x-4=2x-8\)
d) Ta có: \(\dfrac{x^2-5}{x+\sqrt{5}}\)
\(=\dfrac{\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)}{x+\sqrt{5}}\)
\(=x-\sqrt{5}\)
.Cảm ơn!!