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TV
7 tháng 11 2017
a) \(3\sqrt{3}-3\sqrt{4^2\cdot3}+2\sqrt{6^2\cdot3}-\left(2-\sqrt{3}\right)\)
\(3\sqrt{3}-3\cdot4\sqrt{3}+2\cdot6\sqrt{3}-2+\sqrt{3}\)
\(3\sqrt{3}-12\sqrt{3}+12\sqrt{3}-2+\sqrt{3}\)
\(4\sqrt{3}-2\)
b) \(3\sqrt{2}\left(\sqrt{5^2\cdot2}-2\sqrt{3^2\cdot2}+\sqrt{7^2\cdot2}\right)\)
\(3\sqrt{2}\left(5\sqrt{2}-6\sqrt{2}+7\sqrt{2}\right)\)
\(3\sqrt{2}\left(6\sqrt{2}\right)\) \(=36\)
VT
7 tháng 11 2017
\(a=\sqrt{27}-3\sqrt{48}+2\sqrt{108}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(a=3\sqrt{3}-3\sqrt{48}+\sqrt{216}-2+\sqrt{3}\)
\(a=3\sqrt{3}-3\sqrt{48}+3\sqrt{24}-2+\sqrt{3}\)
\(a=3\left(\sqrt{3}-\sqrt{48}+\sqrt{24}+1\right)-2\)
Tính cái trong ngoặc là \(ok\).Em lười đi lấy máy tính lắm
\(b=3\sqrt{2}\left(\sqrt{50}-2\sqrt{18}+\sqrt{98}\right)\)
\(b=3\sqrt{100}-3\sqrt{72}+3\sqrt{196}\)
\(b=3\left(\sqrt{100}-\sqrt{72}+\sqrt{196}\right)\)(Tính trong ngoặc)
ND
27 tháng 7 2018
\(a.3\sqrt{2}\left(\sqrt{50}-2\sqrt{18}+\sqrt{98}\right)=30-36+42=36\)
\(b.B=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{13+30\sqrt{2}+30}=\sqrt{25+2.5\sqrt{18}+18}=\sqrt{\left(5+\sqrt{18}\right)^2}=5+3\sqrt{2}\)
19 tháng 5 2022
a: \(=\sqrt{\dfrac{16}{9}\cdot\dfrac{4}{100}}=\dfrac{4}{3}\cdot\dfrac{2}{10}=\dfrac{4}{3}\cdot\dfrac{1}{5}=\dfrac{4}{15}\)
b: \(=\sqrt{0.09\cdot0.09}\cdot\sqrt{1.21\cdot0.4}\)
\(=0.09\cdot\dfrac{11\sqrt{10}}{50}=\dfrac{99\sqrt{10}}{5000}\)
c: \(=\dfrac{9\sqrt{2}-14\sqrt{2}+6\sqrt{2}}{\sqrt{2}}=9+6-14=1\)
18 tháng 8 2020
a) Ta có: \(D=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\cdot\left(-\sqrt{2}\right)\)
\(=-2+\sqrt{6-2\sqrt{5}}\)
\(=-2+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)
\(=-2+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=-2+\left|\sqrt{5}-1\right|\)
\(=-2+\sqrt{5}-1\)(Vì \(\sqrt{5}>1\))
\(=-3+\sqrt{5}\)
b) Ta có: \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}\right)-\sqrt{75}\)
\(=2\sqrt{81}+4\sqrt{144}-5\sqrt{3}\)
\(=18+48-5\sqrt{3}\)
\(=66-5\sqrt{3}\)
c) Ta có: \(E=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=\sqrt{2}\cdot\left(5-3\right)\)
\(=2\sqrt{2}\)
d) Ta có: \(P=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\frac{3}{2}+2\cdot\sqrt{\frac{3}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}+\sqrt{\frac{3}{2}-2\cdot\sqrt{\frac{3}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}\)
\(=\sqrt{\left(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)^2}\)
\(=\left|\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right|+\left|\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right|\)
\(=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\)(Vì \(\sqrt{\frac{3}{2}}>\sqrt{\frac{1}{2}}>0\))
\(=2\sqrt{\frac{3}{2}}=\sqrt{4\cdot\frac{3}{2}}=\sqrt{6}\)
e) Ta có: \(M=-3\sqrt{50}+2\sqrt{98}-7\sqrt{72}\)
\(=\sqrt{2}\cdot\left(-3\cdot\sqrt{25}+2\cdot\sqrt{49}-7\cdot\sqrt{36}\right)\)
\(=\sqrt{2}\cdot\left(-15+14-42\right)\)
\(=-43\sqrt{2}\)
\(C=3\sqrt{2}\left(\sqrt{50}-2\sqrt{18}+\sqrt{98}\right)\)
\(=3\sqrt{2}.\sqrt{50}-3\sqrt{2}.2\sqrt{18}+3\sqrt{2}.\sqrt{98}\)
\(=3\sqrt{100}-6\sqrt{36}+3\sqrt{196}\)
\(=3.10-6.6+3.14\)
\(=30-36+42\)
\(=36\)