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\(a.\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=\dfrac{2\left(4+3\sqrt{2}\right)-2\left(4-3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{8+6\sqrt{2}-8+6\sqrt{2}}{16-18}=\dfrac{12\sqrt{2}}{-2}=-6\sqrt{2}\)\(b.\dfrac{2}{1+\sqrt{2}}+\dfrac{2}{1-\sqrt{2}}=\dfrac{2\left(1-\sqrt{2}\right)+2\left(1+\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}=\dfrac{2-2\sqrt{2}+2+2\sqrt{2}}{1-2}=-4\)\(c.\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}=14-12\sqrt{7}+18+12\sqrt{7}=14+18=32\)\(d.\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}=14-14\sqrt{2}+7+14\sqrt{2}=14+7=21\)\(e.\left(\sqrt{6}-\sqrt{5}\right)^2-2\sqrt{120}=6-2\sqrt{30}+5-4\sqrt{30}=11-6\sqrt{30}\)
a. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+14\sqrt{2}=14-14\sqrt{2}+7+14\sqrt{2}=21\)
b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{2\sqrt{5}-\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)
c. \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+2\sqrt{7}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
b) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
b) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) = \(1+\sqrt{2}\)
a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\) = \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\) = \(\dfrac{\sqrt{2}}{2}\)
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
a: \(=\left(2\sqrt{7}+\sqrt{7}+2\sqrt{14}\right)\cdot\sqrt{7}-\left(51+14\sqrt{2}\right)\)
\(=3\sqrt{7}\cdot\sqrt{7}+2\sqrt{14}\cdot\sqrt{7}-51-14\sqrt{2}\)
\(=21-51=-30\)
b: \(=\dfrac{\sqrt{10}}{2}+\dfrac{\sqrt{10}-\sqrt{6}}{2}=\dfrac{2\sqrt{10}-\sqrt{6}}{2}\)
c: \(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\sqrt{5}+\sqrt{3}}+\dfrac{\left(\sqrt{5}-\sqrt{2}\right)^2}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{2}\)
\(=2\sqrt{5}+\sqrt{3}-\sqrt{2}\)
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\) = \(\dfrac{\sqrt{2}\sqrt{3}+\sqrt{2}\sqrt{7}}{2\sqrt{3}+2\sqrt{7}}\) = \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\) = \(\dfrac{\sqrt{2}}{2}\)