a: Ta có: \(3\sqrt{2}\cdot5\sqrt{6}\cdot4\sqrt{12}\)

\(=\sqrt{18\cdot25\cdot6\cdot16\cdot12}\)

\(=\sqrt{518400}\)

=720

b: Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)^2+2\sqrt{14}\)

\(=9-2\sqrt{14}+2\sqrt{14}\)

=9

c: Ta có: \(\left(1+\sqrt{5}+\sqrt{6}\right)\left(1+\sqrt{5}-\sqrt{6}\right)\)

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

\(=2\sqrt{5}\)

vt phép tính rõ ra đi

vt là viết

\(2\sqrt{3}\left(\sqrt{12}-\sqrt{27}-3\sqrt{2}\right)+6\sqrt{6}\)

\(=2\sqrt{3}\left(\sqrt{3.2^2}-\sqrt{3.3^2}-3\sqrt{2}\right)+6\sqrt{6}\)

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

\(=\left(2\sqrt{3}\right)^2-6.\left(\sqrt{3}\right)^2-6\sqrt{6}+6\sqrt{6}\)

\(=-6\)

a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)

\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)

\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)

\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)

\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)

\(=-8\sqrt{3}\)

b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)

\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)

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

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

\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

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

\(=2\sqrt{2}\)

\(\sqrt{5}+2\sqrt{6}-\left(\sqrt{5}-2\sqrt{6}\right)=4\sqrt{6}\)

`\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}`

`=\sqrt{2+2.\sqrt{2}.\sqrt{3}+3}-\sqrt{2-2.\sqrt{2}.\sqrt{3}+3}`

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

`=|\sqrt{2}+\sqrt{3}|-|\sqrt{2}-\sqrt{3}|`

`=\sqrt{2}+\sqrt{3}+\sqrt{2}-\sqrt{3}`

`=2\sqrt{2}`

\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

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

\(=2\sqrt{2}\)

\(a_1,\sqrt{x}< 7\\ \Rightarrow x< 49\\ a_2,\sqrt{2x}< 6\\ \Rightarrow x< 18\\ a_3,\sqrt{4x}\ge4\\ \Rightarrow4x\ge16\\ \Rightarrow x\ge4\\ a_4,\sqrt{x}< \sqrt{6}\\ \Rightarrow x< 6\)

\(b_1,\sqrt{x}>4\\ \Rightarrow x>16\\ b_2,\sqrt{2x}\le2\\ \Rightarrow2x\le4\\ \Rightarrow x\le2\\ b_3,\sqrt{3x}\le\sqrt{9}\\ \Rightarrow3x\le9\\ \Rightarrow x\le3\\ b_4,\sqrt{7x}\le\sqrt{35}\\ \Rightarrow7x\le35\\ \Rightarrow x\le5\)

Mình cám ơn Hà Quang Minh rất nhiều

\(\text{a) }Ta\text{ }có:\text{ }\sqrt{5}-\sqrt{3}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{2}{\sqrt{5}+\sqrt{3}}\\ Lại\text{ }có:\text{ }\left(\sqrt{5}+\sqrt{3}\right)^2=5+3+2\sqrt{15}\\ =8+\sqrt{60}< 8+\sqrt{64}=16\\ \Rightarrow\sqrt{5}+\sqrt{3}< 4\\ \Rightarrow\dfrac{2}{\sqrt{5}+\sqrt{3}}>\dfrac{2}{4}\\ \Rightarrow\sqrt{5}-\sqrt{3}>\dfrac{1}{2}\)

\(\text{b) }\sqrt{k+1}-\sqrt{k}=\dfrac{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}}\\ =\dfrac{1}{\sqrt{k+1}+\sqrt{k}}\\ \Rightarrow\sqrt{7}-\sqrt{6}=\dfrac{1}{\sqrt{7}+\sqrt{6}}\\ \sqrt{6}-\sqrt{5}=\dfrac{1}{\sqrt{6}+\sqrt{5}}\\ Mà\text{ }\sqrt{7}+\sqrt{6}>\sqrt{5}+\sqrt{6}\\ \Rightarrow\dfrac{1}{\sqrt{7}+\sqrt{6}}< \dfrac{1}{\sqrt{6}+\sqrt{5}}\\\sqrt{7}-\sqrt{6}< \sqrt{6}-\sqrt{5}\)

Vậy................