\(a,=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\\ b,=\left|1-\sqrt{2}\right|+\sqrt{5}=\sqrt{2}-1+\sqrt{5}\)

a)\(\sqrt{12}+3\sqrt{27}-\sqrt{300}=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\)

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

a: \(A=\sqrt{27}-2\sqrt{48}+3\sqrt{75}\)

\(=3\sqrt3-2\cdot4\sqrt3+3\cdot5\sqrt3\)

\(=3\sqrt3-8\sqrt3+15\sqrt3=18\sqrt3-8\sqrt3=10\sqrt3\)

b: \(B=\sqrt{\left(\sqrt5-2\right)^2}-\sqrt{\left(\sqrt5-3\right)^2}\)

\(=\sqrt5-2-\left|3-\sqrt5\right|\)

\(=\sqrt5-2-3+\sqrt5=-5+2\sqrt5\)

c: \(C=\sqrt{\left(2\sqrt3+1\right)^2}+\sqrt{\left(2\sqrt3-5\right)^2}\)

\(=2\sqrt3+1+\left|5-2\sqrt3\right|\)

\(=2\sqrt3+1+5-2\sqrt3\)

=6

d: \(D=\sqrt{9-4\sqrt5}-\sqrt{14+6\sqrt5}\)

\(=\sqrt{\left(\sqrt5-2\right)^2}-\sqrt{\left(3+\sqrt5\right)^2}\)

\(=\sqrt5-2-\left(3+\sqrt5\right)=\sqrt5-2-3-\sqrt5=-5\)

e: \(E=\frac{4}{\sqrt5-2}-\frac{32}{\sqrt5+1}\)

\(=\frac{4\left(\sqrt5+2\right)}{\left(\sqrt5-2\right)\left(\sqrt5+2\right)}-\frac{32\left(\sqrt5-1\right)}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)

\(=4\left(\sqrt5+2\right)-8\left(\sqrt5-1\right)=4\sqrt5+8-8\sqrt5+8=-4\sqrt5+16\)

f: \(M=\frac{10}{3\sqrt2-4}+\frac{28}{3\sqrt2+2}\)

\(=\frac{10\left(3\sqrt2+4\right)}{\left(3\sqrt2+4\right)\left(3\sqrt2-4\right)}+\frac{28\left(3\sqrt2-2\right)}{\left(3\sqrt2-2\right)\left(3\sqrt2+2\right)}\)

\(=\frac{10\left(3\sqrt2+4\right)}{2}+\frac{28\left(3\sqrt2-2\right)}{18-4}=5\left(3\sqrt2+4\right)+2\left(3\sqrt2-2\right)=15\sqrt2+20+6\sqrt2-4=21\sqrt2+16\)

Giúp mình :<

a) \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=\left(2+6+15-36\right)\sqrt{3}=-13\sqrt{3}\)

b) \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=6\left(3+8-5\right)=36\)

a)\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=\sqrt{4\cdot3}+2\sqrt{9\cdot3}+3\sqrt{25\cdot3}-9\sqrt{16\cdot3}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

b)\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)

\(=2\sqrt{3}\left(\sqrt{9\cdot3}+2\sqrt{16\cdot3}-\sqrt{25\cdot3}\right)\)

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

\(2\sqrt{3}\cdot6\sqrt{3}=12\cdot3=36\)

\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)

\(=9\sqrt{5}-28\sqrt{3}\)

\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)

\(=4-2\cdot5+6-7\)

\(=4-10+6-7\)

=-7

A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)

  =\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)

  =2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)

   =\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)

   =9\(\sqrt{5}\)- 28\(\sqrt{3}\)

a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

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

b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)

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

\(=1-5-2\sqrt{6}\)

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

a) \(-\sqrt{3}\)      b) -10             c)  60               d)  -1             e) 1

a/\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

b/ \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)

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

\(=2\sqrt{3}\cdot6\sqrt{3}=2\cdot6\cdot3=36\)

c/ \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)

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

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

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

d/ \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{8-4\sqrt{10}+5}-\sqrt{45+12\sqrt{10}+8}\)

\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2\cdot5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\cdot2\sqrt{5\cdot2}+\left(2\sqrt{2}\right)^2}\)

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

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

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

#)Giải :

 \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)