a) Ta có: \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)

\(=10\sqrt{5}\)

b) Ta có: \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=3\left(2-\sqrt{3}\right)+4+\sqrt{3}+2\sqrt{3}\)

\(=6-2\sqrt{3}+4+3\sqrt{3}\)

\(=10+\sqrt{3}\)

c) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

=7-5=2

d) Ta có: \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

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

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

a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)

\(=10\sqrt{5}\)

b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\dfrac{6\sqrt{3}}{\sqrt{3}.\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{4-3}+\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+\dfrac{6\sqrt{3}}{3}\)

\(=3\left(2-\sqrt{3}\right)+\dfrac{13\left(4+\sqrt{3}\right)}{13}+2\sqrt{3}\)

\(=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)

\(=10\)

c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=7-5=2\)

d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

\(=\left|2+\sqrt{3}\right|-\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\left|2+\sqrt{3}\right|-\left(5-\sqrt{3}\right)^2\)

\(=\left|2+\sqrt{3}\right|-\left|5-\sqrt{3}\right|\)

\(=2+\sqrt{3}-\left(5-\sqrt{3}\right)\) (vì \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))

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

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

Lời giải:
a.

\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)

b.

\(=\frac{3(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{13(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}+\frac{6\sqrt{3}}{3}\)

\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)

\(=10\)

c.

\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)

\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)

d.

\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)

\(=|2+\sqrt{3}|-|5-\sqrt{3}|=2+\sqrt{3}-(5-\sqrt{3})=-3+2\sqrt{3}\)

a, Ta có : \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(\Rightarrow A^2=2-\sqrt{3}+2+\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

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

\(\Rightarrow A=-\sqrt{2}\)

b, Ta có : \(B=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(\Rightarrow B\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\)

\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)

\(=\sqrt{5}+1+3-\sqrt{5}-2=2\)

\(\Rightarrow B=\sqrt{2}\)

a: A=2-căn 3+căn 3=2

b: B=căn 5+1-căn 5+1=2

a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)

b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)

c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)

d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)

a. \(\sqrt{48}-2\sqrt{32}-\sqrt{75}+3\sqrt{50}\) = \(4\sqrt{3}-2.4\sqrt{2}-5\sqrt{3}+3.5\sqrt{2}\)

= \(4\sqrt{3}-8\sqrt{2}-5\sqrt{3}+15\sqrt{2}\)  = \(-\sqrt{3}+7\sqrt{2}\)

b. \(\sqrt{20}-15\sqrt{\dfrac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}\) = \(2\sqrt{5}-3.5.\sqrt{\dfrac{1}{5}}+\left|1-\sqrt{5}\right|\)

= \(2\sqrt{5}-3\sqrt{25.\dfrac{1}{5}}+\sqrt{5}-1\) =  \(2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1\) = \(-1\)

c. \(\dfrac{3}{3+2\sqrt{3}}+\dfrac{3}{3-2\sqrt{3}}\) = \(\dfrac{3\left(3-2\sqrt{3}\right)+3\left(3+2\sqrt{3}\right)}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\) 

= \(\dfrac{9-6\sqrt{3}+9+6\sqrt{3}}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\) = \(\dfrac{18}{9-12}=\dfrac{18}{-3}=-6\)

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

\(a.\sqrt{72}-5\sqrt{2}+3\sqrt{12}\\ =6\sqrt{2}-5\sqrt{2}+6\sqrt{3}\\ =\sqrt{2}+6\sqrt{3}\\ b.6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\\ =3\sqrt{2}-\sqrt{2}-5\sqrt{2}\\ =-3\sqrt{2}\\ c.\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\\ =2+1+\sqrt{3}-\sqrt{3}\\ =3\\ d.\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\\ =4+3+4\\ =11\)

\(A=2\sqrt{27}-\sqrt{\left(1-\sqrt{3}\right)^2}+\dfrac{1}{2-\sqrt{3}}\\ =2.3\sqrt{3}-\left|1-\sqrt{3}\right|+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ =6\sqrt{3}-\left(-1+\sqrt{3}\right)+\dfrac{2+\sqrt{3}}{2^2-\sqrt{3^2}}\\ =6\sqrt{3}+1-\sqrt{3}+2+\sqrt{3}\\ =6\sqrt{3}+3\)

\(=6\sqrt{3}-\sqrt{3}+1+2+\sqrt{3}=6\sqrt{3}+3\)