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\(\sqrt{3\cdot27}-\sqrt{\dfrac{144}{36}}\)=\(\sqrt{81}-\sqrt{4}\)=9-2=7
\(\dfrac{2\cdot3+3\cdot6}{4}\)=6
\(\sqrt{7}-\sqrt{7-2\cdot\sqrt{7}+1}\)=\(\sqrt{7}-\left(\sqrt{7}-1\right)\)=1
\(\dfrac{\sqrt{3-2\cdot\sqrt{3}+1}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=\(\dfrac{\sqrt{3}-1}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=\(\dfrac{1}{\sqrt{2}}\)
\(\dfrac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)+\(\dfrac{\sqrt{3}\cdot\left(1+\sqrt{3}\right)}{\sqrt{3}+1}\)-(\(\sqrt{5}+3\))
=(\(\sqrt{5}+3\))+\(\sqrt{3}\)-(\(\sqrt{5}+3\))=\(\sqrt{3}\)
\(\sqrt{3}\cdot\sqrt{9}+5\cdot\sqrt{4}\cdot3-2\sqrt{3}\)
=\(\sqrt{3}\cdot\left(3+10-2\right)\)
=\(11\sqrt{3}\)
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
1.\(=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}=3\sqrt{2}\)
2.\(=5\sqrt{5}+4\sqrt{5}-9\sqrt{5}=0\)
3: \(\sqrt{12-3\sqrt{7}}-\sqrt{12-3\sqrt{7}}=0\)
4: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)
\(=-2\sqrt{2}\)
6: \(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
\(=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)
\(=-4\sqrt{3}\)
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
a: \(=\sqrt{11}-1\)
b: \(=3\sqrt{3}+1\)
c: \(=\sqrt{3}+\sqrt{2}\)
d: \(=\sqrt{3}-\sqrt{2}\)
e: \(=\sqrt{3}-1\)
g: \(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
a,\(\sqrt{2\left(11+6\sqrt{2}\right)}\)=\(\sqrt{2\left(9+2.3.\sqrt{2}+2\right)}\)=\(\sqrt{2\left(3+\sqrt{2}\right)^2}\)=\(\sqrt{2}\)(3+\(\sqrt{2}\))
\(a.\sqrt{22+12\sqrt{2}}=\sqrt{18+2.3\sqrt{2}.2+4}=3\sqrt{2}+2\)
\(b.\sqrt{\dfrac{5+2\sqrt{6}}{2}}=\sqrt{\dfrac{3+2\sqrt{3}.\sqrt{2}+2}{2}}=\dfrac{\sqrt{3}+\sqrt{2}}{2}\)
\(c.\sqrt{30+4\sqrt{2}.\sqrt{7}}=\sqrt{28+2.\sqrt{2}.2\sqrt{7}+2}=2\sqrt{7}+\sqrt{2}\)
\(d.\sqrt{5+2\sqrt{2-\sqrt{9-4\sqrt{2}}}}=\sqrt{5+2\sqrt{2-\sqrt{8-2.2\sqrt{2}+1}}}=\sqrt{5+2\sqrt{2-2\sqrt{2}+1}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{2}+1\) \(e.\sqrt{1+2\sqrt{\sqrt{2+\sqrt{11+6\sqrt{2}}}}}=\sqrt{1+2\sqrt{\sqrt{2+\sqrt{9+2.3\sqrt{2}+2}}}}=\sqrt{1+2\sqrt{\sqrt{5+\sqrt{2}}}}\)
\(f.\sqrt{1+\dfrac{\sqrt{3}}{2}+\sqrt{1-\dfrac{\sqrt{3}}{2}}}=\sqrt{1+\dfrac{\sqrt{3}}{2}+\sqrt{\dfrac{3}{4}-2.\dfrac{\sqrt{3}}{2}.\dfrac{1}{2}+\dfrac{1}{4}}}=\sqrt{\sqrt{3}+\dfrac{1}{2}}=\)
\(g.\sqrt{10-2\sqrt{21}}+\sqrt{4+2\sqrt{3}}=\sqrt{7-2\sqrt{7}.\sqrt{3}+3}+\sqrt{3+2\sqrt{3}+1}=\sqrt{7}-\sqrt{3}+\sqrt{3}+1=\sqrt{7}+1\)
b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
= \(\sqrt{3.4-3\sqrt{7}}-\sqrt{3.4+3\sqrt{7}}\)
= \(\sqrt{3.\left(4-\sqrt{7}\right)}-\sqrt{3.\left(4+\sqrt{7}\right)}\)
= \(\sqrt{3}.\sqrt{4-\sqrt{7}}-\sqrt{3}.\sqrt{4+\sqrt{7}}\)
= \(\sqrt{3}.\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)\)
\(\)≈ \(-2,449\)
\(\sqrt{\dfrac{13}{4}+\sqrt{3}}-\sqrt{\dfrac{7}{4}-\sqrt{3}}\)
= \(\sqrt{\dfrac{13}{4}+\dfrac{4\sqrt{3}}{4}}-\sqrt{\dfrac{7}{4}-\dfrac{4\sqrt{3}}{4}}\)
= \(\sqrt{\dfrac{13+4\sqrt{3}}{4}}-\sqrt{\dfrac{7-4\sqrt{3}}{4}}\)
= \(\dfrac{\sqrt{13+4\sqrt{3}}}{\sqrt{4}}-\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)
= \(\dfrac{\sqrt{13+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)
≈ \(2,098\)
\(7\sqrt{12}+6\sqrt{27}=7\cdot2\sqrt{3}+6\cdot3\sqrt{3}=14\sqrt{3}+18\sqrt{3}=32\sqrt{3}\\ \dfrac{2}{\sqrt{11}-3}-\sqrt{11}=\dfrac{2\left(\sqrt{11}+3\right)}{11-9}-\sqrt{11}\\ =\dfrac{2\left(\sqrt{11}+3\right)}{2}-\sqrt{11}=\sqrt{11}+3-\sqrt{11}=3\)