\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)

\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)

\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)

Giải chi tiết từng bước nha

Câu b á bạn, chỗ \(\dfrac{4}{\sqrt{5-1}}\) là đề như vậy hay là \(\dfrac{4}{\sqrt{5}-1}\) vậy?

\(\text{Theo đề bài: }=\dfrac{3\sqrt{2}+6\sqrt{3}+2\sqrt{5}-\sqrt{6}}{2}\)

6 + 2 căn 2 + 2 căn 3 + 2 căn 6 = (1 + căn 2 + căn 3)² 

5 + 2 căn 6 = (căn 2 + căn 3)²

tự giải tieeps nha bn ^^

+) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)

\(=3\sqrt{4.5}-2\sqrt{9.5}+4\sqrt{5}\)

\(=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)

\(=4\sqrt{5}\)

+) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=\left(2\sqrt{7}-\sqrt{28}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=\left(2\sqrt{7}-2\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=7+7\sqrt{8}\)