\(\sqrt{28}+2\sqrt{75}\)
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17 tháng 12 2023
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: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)
\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
\(=-16\sqrt{3}\)
25 tháng 8 2018
\(A=\sqrt{75}-3\sqrt{63}-2\sqrt{27}+5\sqrt{28}\)
\(=5\sqrt{3}-9\sqrt{7}-6\sqrt{3}+10\sqrt{7}\)
\(=\sqrt{7}-\sqrt{3}\)
\(B=\sqrt{\left(3+2\sqrt{5}\right)^2}-\sqrt{\left(3-2\sqrt{5}\right)^2}\)
\(=3+2\sqrt{5}-\left|3-2\sqrt{5}\right|\)
\(=3+2\sqrt{5}-2\sqrt{5}+3\)
\(=6\)
ND
17 tháng 7 2018
\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=\left(6-4+50-63\right)\sqrt{3}=-11\sqrt{3}\)
\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=\left(2+9-30-\dfrac{6}{5}\right)\sqrt{7}=-20,2\sqrt{7}\)\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=10\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-12\sqrt{11}\)
\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)
17 tháng 7 2018
\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=-11\sqrt{3}\)
\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=-\dfrac{101}{5}\sqrt{7}\)
\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=20\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-2\sqrt{11}\)
\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)
4 tháng 7 2018
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
17 tháng 7 2017
a, \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
= \(2\sqrt{3}-10\sqrt{3}-\dfrac{\sqrt{3}\cdot\sqrt{11}}{\sqrt{11}}+5\sqrt{\dfrac{4}{3}}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\sqrt{\dfrac{12}{3^2}}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\dfrac{2\sqrt{3}}{3}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}\)
= \(-9\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{-27\sqrt{3}}{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{-17\sqrt{3}}{3}\)
b, \(\sqrt{150}+\sqrt{1,6}\cdot\sqrt{60}+4.5\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
= \(5\sqrt{6}+\dfrac{2\sqrt{10}}{5}\cdot2\sqrt{15}+4,5\sqrt{\dfrac{8}{3}}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+4,5\sqrt{\dfrac{24}{3^2}}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+4,5\cdot\dfrac{2\sqrt{6}}{3}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)
17 tháng 7 2017
c, \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\)
= \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
d, \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{30}+5-2\sqrt{30}=11\)