a)

\(\sqrt{12}-\sqrt{27}+\sqrt{3}=\sqrt{4}.\sqrt{3}-\sqrt{9}.\sqrt{3}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}\)

\(=\sqrt{3}(2-3+1)=0\)

b)

\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}.\sqrt{63}-\sqrt{4}.\sqrt{175}+\sqrt{4}.\sqrt{252}-\sqrt{4}.\sqrt{112}\)

\(=2(\sqrt{63}-\sqrt{175}+\sqrt{252}-\sqrt{112})\)

\(=2(\sqrt{9}.\sqrt{7}-\sqrt{25}.\sqrt{7}+\sqrt{36}.\sqrt{7}-\sqrt{16}.\sqrt{7})\)

\(=2(3\sqrt{7}-5\sqrt{7}+6\sqrt{7}-4\sqrt{7})=2\sqrt{7}(3-5+6-4)=0\)

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\(\sqrt{3}(\sqrt{12}+\sqrt{27}-\sqrt{3})=\sqrt{36}+\sqrt{81}-\sqrt{9}\)

\(=\sqrt{6^2}+\sqrt{9^2}-\sqrt{3^2}=6+9-3=12\)

c)

\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\frac{\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{5}}{\sqrt{7}.\sqrt{3}+\sqrt{7}.\sqrt{5}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{5})}{\sqrt{7}(\sqrt{3}+\sqrt{5})}=\frac{\sqrt{2}}{\sqrt{7}}\)

\(\frac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\frac{\sqrt{81}.\sqrt{5}+3\sqrt{9}.\sqrt{3}}{3\sqrt{3}+\sqrt{9}.\sqrt{5}}=\frac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}\)

\(=\frac{3(3\sqrt{5}+3\sqrt{3})}{3\sqrt{3}+3\sqrt{5}}=3\)

d)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{6}+\sqrt{9}+\sqrt{12})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

\(1,=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\\ 2,=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right):5=\dfrac{2\sqrt{6}}{5}-\dfrac{2\sqrt{5}}{5}\\ 3,=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-\dfrac{9\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\\ 4,Sửa:\dfrac{1}{\sqrt{5}-\sqrt{3}}-\dfrac{1}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)

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

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

3) \(=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)

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

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, \(A=2\sqrt{3}-\sqrt{12}-\sqrt{9}\)

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

b, \(B=\sqrt{3}\left(\sqrt{12}+\sqrt{27}\right)\)

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

\(=\sqrt{3}.5\sqrt{3}=5.3=15\)

\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

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

\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)

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

\(=-5\)

\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)

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

\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)

\(=\sqrt{5}.\left(5-12+6-4\right)\)

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

\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)

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

\(=2.3+5-5.2\)

\(=1\)

\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)

\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)

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

\(=0\)

\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)

\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)

\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)

\(=7\sqrt{3}\)

\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)

\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)

L=2*3+5-5*2=5-4=1

N=8căn 2-8căn2-15căn3+15căn 3=0

O=10căn 3-6căn3+3căn3=7căn 3

Bấm máy tính là ra thui mà bn

a/ \(=2\sqrt{3}-3\sqrt{3}+\sqrt{3}=0\)

b/ \(=\left(2\sqrt{3}-10\sqrt{3}\right)\sqrt{3}=-24\)

c/ \(=15-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=15-6\sqrt{7}\)

d/ \(=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\right)=12\)

f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)

mik sửa lại câu f , tí nhé :

f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)

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

3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\dfrac{3}{2}\sqrt{12}\right)\)

=\(\sqrt{3}\left(2.3\sqrt{3}-5\sqrt{3}+\dfrac{3}{2}.2\sqrt{3}\right)\)

=\(\sqrt{3}\left(6\sqrt{3}-5\sqrt{3}+3\sqrt{3}\right)\)

=\(\sqrt{3}.\left(4\sqrt{3}\right)\)

=12

\(\left(2\sqrt{3}+2.3\sqrt{3}-\sqrt{3}\right):\sqrt{3}\)

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

=\(\left(7\sqrt{3}\right):\sqrt{3}\)

=\(7\sqrt{3}.\dfrac{1}{\sqrt{3}}\)

=7

a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)

c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)

d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)

e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)

f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)

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

\(=\sqrt{12}:\sqrt{3}+2\sqrt{27}:\sqrt{3}-\sqrt{3}:\sqrt{3}\)

\(=\sqrt{4}+2\sqrt{9}-1\)

\(=2+6-1\)

\(=7\)

2) \(\left(4\sqrt{2}-\sqrt{8}+2\right).\sqrt{2-\sqrt{8}}\)

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

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

\(=\left(8+4\right)\left(2-2\sqrt{2}\right)\)

\(=12.\left(2-2\sqrt{2}\right)\)

\(=24-24\sqrt{2}\)

\(=24\left(1-\sqrt{2}\right)\)

3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\frac{3}{2}\sqrt{12}\right)\)

\(=\sqrt{3}\left(2\sqrt{3^2.3}-\sqrt{5^2.3}+\frac{3}{2}\sqrt{2^2.3}\right)\)

\(=\sqrt{3}\left(6\sqrt{3}-5\sqrt{3}+3\sqrt{3}\right)\)

\(=\sqrt{3}.4\sqrt{3}\)

\(=12\)