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1. \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+\sqrt{84}\)= -6,423305878
2. \(\sqrt{150}+\sqrt{1,6}\sqrt{60}+4,5\sqrt{2\frac{2}{3}}-\sqrt{6}\)= 24,79207036
NHA 
Vũ Hoàng Thiên An ! ! !
K VÀ KB NHA !
1) Ta có: \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=1+\sqrt{2}\)
2) Ta có: \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
\(=\sqrt{108}-\sqrt{36\cdot\frac{4}{3}}+\sqrt{75\cdot\frac{9}{25}}\)
\(=\sqrt{108}-\sqrt{48}+\sqrt{27}\)
\(=\sqrt{3}\left(6-4+3\right)\)
\(=5\sqrt{3}\)
3) Sửa đề: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
Ta có: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
\(=\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{3}-10\sqrt{4}\cdot\sqrt{3}+16\cdot\sqrt{4}\cdot\sqrt{3}\)
\(=\sqrt{2}\cdot\sqrt{12}-10\sqrt{12}+16\sqrt{12}\)
\(=\sqrt{12}\left(\sqrt{2}-10+16\right)\)
\(=2\sqrt{3}\left(\sqrt{2}-6\right)\)
\(=2\sqrt{6}-12\sqrt{3}\)
4) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{\sqrt{12}}{6}-\frac{2\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6\left(2-\sqrt{3}\right)+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{12-6\sqrt{3}+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{6-2\sqrt{3}}{6}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2\sqrt{3}\cdot\sqrt{3}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{3}}\)
5) Ta có: \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\frac{\sqrt{3}\left(2+5+3\right)}{\sqrt{15}}=\frac{10}{\sqrt{5}}=2\sqrt{5}\)
6) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48\cdot\frac{1}{4}}-\sqrt{75\cdot4}-\sqrt{3}+5\sqrt{\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{25\cdot\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{\frac{100}{3}}\)
\(=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)\)
\(=-\frac{17\sqrt{3}}{3}=-\frac{17}{\sqrt{3}}\)
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, \(\sqrt{150}+\sqrt{1,6}\cdot\sqrt{60}+4.5\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
= \(5\sqrt{6}+\dfrac{2\sqrt{10}}{5}\cdot2\sqrt{15}+4,5\sqrt{\dfrac{8}{3}}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+4,5\sqrt{\dfrac{24}{3^2}}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+4,5\cdot\dfrac{2\sqrt{6}}{3}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)
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, \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{30}+5-2\sqrt{30}=11\)
1 a/ Trục căn thức ở mẫu
\(VT=\frac{-\sqrt{1}+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{47}+\sqrt{48}}{48-47}\)\(=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{47}+\sqrt{48}=\sqrt{48}-1>3=VP\)
b/
\(2\left(10+3\sqrt{11}\right)=11+2.\sqrt{11}.3+9=\left(\sqrt{11}+3\right)^2\)
\(VT=\left(\sqrt{11}-3\right)\sqrt{2}\sqrt{10+3\sqrt{11}}=\left(\sqrt{11}-3\right)\left(\sqrt{11}+3\right)=11-9=2=VP\)
2/
\(B=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{2\left(5+\sqrt{3}.\sqrt{7}\right)}\)
\(2\left(5+\sqrt{21}\right)=7+2\sqrt{7}.\sqrt{3}+3=\left(\sqrt{7}+\sqrt{3}\right)^2\)
\(B=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)=\left(5+\sqrt{21}\right).4\)
\(=20+4\sqrt{21}\)
A chắc không rút gọn được.
\(A=\left(\sqrt{8}-3\sqrt{2}+10\right)\left(\sqrt{2}-3\sqrt{0.4}\right)=\sqrt{16}-\frac{12\sqrt{5}}{5}+\sqrt{20}-6\sqrt{10}-6+\frac{18\sqrt{5}}{5}\)
\(A=-2+\frac{16\sqrt{5}}{5}-6\sqrt{10}\)
b)\(B=\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{6+2\sqrt{5}}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{5}+1}{2}-\frac{\sqrt{5}-1}{2}=1\)
\(=\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+\frac{10}{\sqrt{3}}=-9\sqrt{3}+\frac{10}{\sqrt{3}}=\frac{-17\sqrt{3}}{3}\)
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
=\(\frac{1}{2}\sqrt{3.4^2}-2\sqrt{3.5^2}-\sqrt{\frac{33}{11}}+5\sqrt{\frac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+10\sqrt{\frac{1}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10}{3}\sqrt{3}\)
\(=\left(2-10-1+\frac{10}{3}\right)\sqrt{3}\)
\(=\frac{-17}{3}\sqrt{3}\)
\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5\sqrt{2\frac{2}{3}}-\sqrt{6}\)
\(=\sqrt{6.5^2}+\sqrt{96}+4,5\sqrt{\frac{8}{3}}-\sqrt{6}\)
\(=5\sqrt{6}+\sqrt{6.4^2}+4,5\frac{\sqrt{24}}{3}-\sqrt{6}\)
\(=5\sqrt{6}+4\sqrt{6}+\frac{4,5.2\sqrt{6}}{3}-\sqrt{6}\)
\(=8\sqrt{6}+3\sqrt{6}\)
\(=11\sqrt{6}\)
Tự hòi tự trl :D ?
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}\sqrt{16.3}-2.5\sqrt{3}-\sqrt{3}-\frac{10}{3}\sqrt{3}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}-\frac{10}{3}\sqrt{3}\)
\(=-9\sqrt{3}+\frac{10}{3}\sqrt{3}=\left(-9+\frac{10}{3}\right)\sqrt{3}\)
\(=-\frac{17}{3}\sqrt{3}\)
\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\frac{2}{3}}-\sqrt{6}\)
\(=\sqrt{25.6}+\sqrt{1,6.60}+4,8\sqrt{\frac{8}{3}}-\sqrt{6}\)
\(=5\sqrt{6}+\sqrt{16.6}+4,5.\frac{1}{3}\sqrt{3^2.\frac{4.2}{3}}-\sqrt{6}\)
\(=9\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)