a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

        \(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

        \(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)

        \(=19\sqrt{3\sqrt{2}}\)

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

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

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

b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)

= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)

= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)

= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)

a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)

\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)

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

=0

b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)

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

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

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

=2

c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)

\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)

\(=4-\sqrt{7}+\sqrt{7}\)

=4

d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)

\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)

\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)

\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)

\(=16\sqrt{5}\)

e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)

\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)

\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)

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

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

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

\(A=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{2}.\sqrt{6-2\sqrt{5}}+\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}}{2\left(\sqrt{2}+1\right)}\)

\(=\dfrac{\sqrt{5}+1-\sqrt{2}\left(\sqrt{5}-1\right)+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)

\(=\dfrac{\sqrt{5}+1-\sqrt{10}+\sqrt{2}+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)

\(=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}=\dfrac{1}{2}\)

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

\((15\sqrt{200}-3\sqrt{450}+2\sqrt{50}):\sqrt{10}=\left(15.10\sqrt{2}-3.15\sqrt{2}+2.5\sqrt{2}\right):\sqrt{10}=\frac{115\sqrt{2}.1}{\sqrt{10}}=\frac{115\sqrt{20}}{10}\)

Hạ Hy bạn xem lại câu b đúng đề chưa

a)\(\left(\sqrt{10}-\sqrt{15}+3\sqrt{3}\right)\sqrt{5}-\sqrt{72}\)

\(=\sqrt{15}-\sqrt{15}+15-6\sqrt{2}\)

\(15-6\sqrt{2}\)

b)\(\dfrac{\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right)}{8\sqrt{10}}\)

\(=\dfrac{\left(15.5\sqrt{2}+5.10\sqrt{2}-3.15\sqrt{2}\right)}{8\sqrt{10}}\)

\(=\dfrac{\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right)}{8\sqrt{10}}\)

\(=\dfrac{80\sqrt{2}}{8\sqrt{10}}=\dfrac{10\sqrt{2}}{\sqrt{10}}=\sqrt{20}=2\sqrt{5}\)

Giải:

a) \(\left(\sqrt{3}+2\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)

\(=\sqrt{3}.\sqrt{3}+2\sqrt{5}.\sqrt{3}-\sqrt{60}\)

\(=3+2\sqrt{15}-\sqrt{60}\)

\(=3+2\sqrt{15}-2\sqrt{15}\)

\(=3\)

Vậy ...

b) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)

\(=\left(15\sqrt{4.50}-3\sqrt{9.50}+2\sqrt{50}\right):\sqrt{10}\)

\(=\left(30\sqrt{50}-9\sqrt{50}+2\sqrt{50}\right):\sqrt{10}\)

\(=23\sqrt{50}:\sqrt{10}\)

\(=\dfrac{23\sqrt{50}}{\sqrt{10}}\)

\(=\dfrac{23\sqrt{5}\sqrt{10}}{\sqrt{10}}\)

\(=23\sqrt{5}\)

Vậy ...

\(a\text{) }\left(\sqrt{3}+2\sqrt{5}\right)\sqrt{3}-\sqrt{60}\\ =3+2\sqrt{15}-2\sqrt{15}=3\)

\(b\text{) }\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\\ =15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\\ =30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\\ =\left(30-9+2\right)\sqrt{5}=23\sqrt{5}\)