\(a,\sqrt{2^3}=2^{\dfrac{3}{2}}\\ b,\sqrt[5]{\dfrac{1}{27}}=\sqrt[5]{3^{-3}}=3^{-\dfrac{3}{5}}\\ c,\left(\sqrt[5]{a}\right)^4=\sqrt[5]{a^4}=a^{\dfrac{4}{5}}\)

a: \(=3\cdot3^{\dfrac{1}{2}}\cdot3^{\dfrac{1}{.4}}\cdot3^{\dfrac{1}{8}}=3^{1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}}=3^{\dfrac{15}{16}}\)

b: \(=\sqrt{a\cdot\sqrt{a\cdot a^{\dfrac{1}{2}}}}\)

\(=\sqrt{a\cdot\sqrt{a^{\dfrac{3}{2}}}}=\sqrt{a\cdot a^{\dfrac{3}{4}}}=\sqrt{a^{\dfrac{7}{4}}}=a^{\dfrac{7}{4}\cdot\dfrac{1.}{2}}=a^{\dfrac{7}{8}}\)

c: \(=\dfrac{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}}}{\left(a^{\dfrac{1}{5}}\right)^3\cdot a^{\dfrac{2}{5}}}=\dfrac{a^{\dfrac{13}{12}}}{a}=a^{\dfrac{1}{12}}\)

\(a,a^{\dfrac{3}{5}}\cdot a^{\dfrac{1}{2}}:a^{-\dfrac{2}{5}}=a^{\dfrac{3}{5}+\dfrac{1}{2}-\left(-\dfrac{2}{5}\right)}=a^{\dfrac{3}{2}}\\ b,\sqrt{a^{\dfrac{1}{2}}\sqrt{a^{\dfrac{1}{2}}\sqrt{a}}}\\ =\sqrt{a^{\dfrac{1}{2}}\sqrt{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}}}}\\ =\sqrt{a^{\dfrac{1}{2}}\sqrt{a}}\\ =\sqrt{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}}}\\ =\sqrt{a}\)

G/s   \(2015+2014\sqrt{3}=\left(a+b\sqrt{3}\right)^2=a^2+3b^2+2\sqrt{3}ab\)

\(2014\sqrt{3}-2\sqrt{3}ab=a^2+3b^2-2015\)

\(\sqrt{3}\left(2014-2ab\right)=a^2+3b^2-2015\)

\(\sqrt{3}=\frac{a^2+3b^2-2015}{2014-2ab}\)

Với a; b nguyên  =>VP nguyên

mà VT là số vô tỉ 

=> g/s sai 

Vây  

4. \(\sqrt{x}+\sqrt{y}=6\sqrt{55}\)

\(6\sqrt{55}\)  là số vô tỉ, suy ra vế trái phải là các căn thức đồng dạng chứa  \(\sqrt{55}\)

Đặt  \(\sqrt{x}=a\sqrt{55};\sqrt{y}=b\sqrt{55}\)  với  \(a,b\in N\)

\(\Rightarrow a+b=6\)

Xét các TH:

a = 0 => b = 6

a = 1 => b = 5

a = 2 => b = 4

a = 3 => b = 3

a = 4 => b = 2

a = 5 => b = 1

a = 6 => b = 0

Từ đó dễ dàng tìm đc x, y

Biên cưng. Minh Quân đây. 

\(13-4\sqrt{3}=\left(2\sqrt{3}\right)^2-2.2\sqrt{2}.1+1^2=\left(2\sqrt{3}-1\right)^2\)

a) \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)

câu này \(\sqrt{15}\)đúng hơn \(\sqrt{5}\)

b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)c) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)

\(\sqrt{\dfrac{3+\sqrt{5}}{2}}=\sqrt{\dfrac{6+2\sqrt{5}}{2}}=\sqrt{\dfrac{\left(1+\sqrt{5}\right)^2}{2}=\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}}\)

\(\sqrt{\dfrac{3+\sqrt{5}}{2}}=\sqrt{\dfrac{6+2\sqrt{5}}{4}}=\dfrac{\sqrt{5}+1}{2}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\sqrt{5}\)