Ta có: \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)

\(=14+18-6\sqrt{28}+6\sqrt{28}\)

=32

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

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

ta có 

\(\frac{2.6.10+6.10.14+..+194.198.202}{1.3.5+3.5.7+..+97.99.101}=\frac{8.1.3.5+8.3.5.7+..+8.97.99.101}{1.3.5+3.5.7+..+97.99.101}\)

\(=\frac{8.\left(1.3.5+3.5.7+..+97.99.101\right)}{1.3.5+3.5.7+..+97.99.101}=8\)

a: \(=3\cdot7\sqrt{3}+2\cdot6\sqrt{3}-4\cdot4\sqrt{3}-11\sqrt{3}\)

\(=21\sqrt{3}+12\sqrt{3}-16\sqrt{3}-11\sqrt{3}\)

\(=6\sqrt{3}\)

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

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

=2

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

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

=16-3

=13

a) Ta có: \(\dfrac{7\cdot25}{14\cdot10}\)

\(=\dfrac{7\cdot5\cdot5}{7\cdot2\cdot2\cdot5}\)

\(=\dfrac{5}{4}\)

b) Ta có: \(\dfrac{24\cdot15-14\cdot9}{36\cdot12}\)

\(=\dfrac{9\cdot8\cdot5-14\cdot9}{36\cdot12}\)

\(=\dfrac{9\cdot\left(8\cdot5-14\right)}{9\cdot4\cdot12}\)

\(=\dfrac{40-14}{4\cdot12}\)

\(=\dfrac{13}{24}\)

a) \(\dfrac{7.25}{14.10}\)

= \(\dfrac{7.5.5}{7.2.5.2}\)

= \(\dfrac{5}{4}\)

c) \(\dfrac{1^3.3^2}{6^3.5}\)

= \(\dfrac{3^2}{\left(2.3\right)^3.5}\)

= \(\dfrac{1}{2^3.3.5}\)

= \(\dfrac{1}{120}\)

a) \(\dfrac{5}{6}-\dfrac{2}{14}\)

\(=\dfrac{5}{6}-\dfrac{1}{7}\)

\(=\dfrac{35}{42}-\dfrac{6}{42}\)

\(=\dfrac{29}{42}\)

b) \(\dfrac{5}{20}-\dfrac{1}{6}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}\)

\(=\dfrac{3}{12}-\dfrac{2}{12}\)

\(=\dfrac{1}{12}\)

c) \(\dfrac{5}{9}-\dfrac{3}{12}\)

\(=\dfrac{5}{9}-\dfrac{1}{4}\)

\(=\dfrac{20}{36}-\dfrac{9}{36}\)

\(=\dfrac{11}{36}\)

d) \(8-\dfrac{4}{6}\)

\(=\dfrac{48}{6}-\dfrac{4}{6}\)

\(=\dfrac{44}{6}=\dfrac{22}{3}\).

`a)(\sqrt{14}-3\sqrt{2})^2+6\sqrt{28}`

`=14-12\sqrt{7}+18+12\sqrt{7}=32`

`b)2\sqrt{20}-3\sqrt{20}+\sqrt{125}`

`=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}`

`=3\sqrt{5}`.

a) \(\left(\sqrt{14}-3\sqrt{2}\right)^2-6\sqrt{28}\)

\(=\left(\sqrt{14}\right)^2-2\cdot\sqrt{14}\cdot3\sqrt{2}+\left(3\sqrt{2}\right)^2+6\sqrt{28}\)

\(=14-6\sqrt{28}+18+6\sqrt{28}\)

\(=14+18\)

\(=32\)

b) \(2\sqrt{20}-3\sqrt{20}+\sqrt{125}\)

\(=2\cdot2\sqrt{5}-3\cdot2\sqrt{5}+5\sqrt{5}\)

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

\(=3\sqrt{5}\)

a) \(\dfrac{2}{3}-\dfrac{2}{6}=\dfrac{4}{6}-\dfrac{2}{6}=\dfrac{4-2}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)

b) \(\dfrac{5}{6}-\dfrac{3}{18}=\dfrac{10}{18}-\dfrac{3}{18}=\dfrac{10-3}{18}=\dfrac{7}{18}\)

c) \(\dfrac{8}{14}-\dfrac{2}{7}=\dfrac{8}{14}-\dfrac{4}{14}=\dfrac{8-4}{14}=\dfrac{4}{14}=\dfrac{2}{7}\)

d) \(\dfrac{12}{20}-\dfrac{2}{5}=\dfrac{12}{20}-\dfrac{8}{20}=\dfrac{12-8}{20}=\dfrac{4}{20}=\dfrac{1}{5}\)