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a/ \(\sqrt{45}\).\(\sqrt{15}\).\(\sqrt{27}\)
b/\(\frac{\sqrt{27}}{\sqrt{12}}\)+ \(\frac{1}{2}\)
c/\(\sqrt{\frac{1}{3}}\): \(\sqrt{\frac{27}{50}}\). \(\sqrt{2}\)
d/ \(\sqrt{117^2-108^2}\)
K
Khách
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Những câu hỏi liên quan
25 tháng 6 2019
a) \(\sqrt{45}\cdot\sqrt{15}\cdot\sqrt{27}=\sqrt{45\cdot15\cdot27}=135\)
b) \(\frac{\sqrt{17}}{\sqrt{12}}+\frac{1}{2}=\frac{\sqrt{51}}{6}+\frac{3}{6}=\frac{\sqrt{51}+3}{6}\)
c) \(\sqrt{\frac{1}{3}}:\sqrt{\frac{27}{50}}\cdot\sqrt{2}=\sqrt{\frac{1}{3}\cdot\frac{50}{27}\cdot2}=\frac{10}{9}\)
d) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9\cdot225}=45\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 10 2019
\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)
Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3
\(B=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
\(C=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{5}-\sqrt{3}\right)\left(-\sqrt{5}-\sqrt{3}\right)=1+\left(5-3\right)=3\)
\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)
12 tháng 10 2019
a) \(\left(\sqrt{27}-\sqrt{12}-\sqrt{108}-\sqrt{192}\right):\sqrt{3}=\left(3\sqrt{3}-2\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):\sqrt{3}=\left(-13\sqrt{3}\right):\sqrt{3}=-13\sqrt{3}.\frac{1}{\sqrt{3}}=-13\)
c) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
VH
12 tháng 10 2019
a, \(=\left(3\sqrt{3}-2\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):\sqrt{3}\)
\(=\frac{-13\sqrt{3}}{\sqrt{3}}=-13\)
b, \(=\frac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}.\frac{3+2\sqrt{7}}{1+\sqrt{3}}\)
\(=\frac{\sqrt{2}\left(3+2\sqrt{7}\right)}{1+\sqrt{3}}\)
c, \(=\sqrt{6-6\sqrt{6} +9}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
AH
Akai Haruma
Giáo viên
30 tháng 8 2019
a)
\(\sqrt{12}-\sqrt{27}+\sqrt{3}=\sqrt{4}.\sqrt{3}-\sqrt{9}.\sqrt{3}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}\)
\(=\sqrt{3}(2-3+1)=0\)
b)
\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}.\sqrt{63}-\sqrt{4}.\sqrt{175}+\sqrt{4}.\sqrt{252}-\sqrt{4}.\sqrt{112}\)
\(=2(\sqrt{63}-\sqrt{175}+\sqrt{252}-\sqrt{112})\)
\(=2(\sqrt{9}.\sqrt{7}-\sqrt{25}.\sqrt{7}+\sqrt{36}.\sqrt{7}-\sqrt{16}.\sqrt{7})\)
\(=2(3\sqrt{7}-5\sqrt{7}+6\sqrt{7}-4\sqrt{7})=2\sqrt{7}(3-5+6-4)=0\)
------------------
\(\sqrt{3}(\sqrt{12}+\sqrt{27}-\sqrt{3})=\sqrt{36}+\sqrt{81}-\sqrt{9}\)
\(=\sqrt{6^2}+\sqrt{9^2}-\sqrt{3^2}=6+9-3=12\)
AH
Akai Haruma
Giáo viên
30 tháng 8 2019
c)
\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\frac{\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{5}}{\sqrt{7}.\sqrt{3}+\sqrt{7}.\sqrt{5}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{5})}{\sqrt{7}(\sqrt{3}+\sqrt{5})}=\frac{\sqrt{2}}{\sqrt{7}}\)
\(\frac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\frac{\sqrt{81}.\sqrt{5}+3\sqrt{9}.\sqrt{3}}{3\sqrt{3}+\sqrt{9}.\sqrt{5}}=\frac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}\)
\(=\frac{3(3\sqrt{5}+3\sqrt{3})}{3\sqrt{3}+3\sqrt{5}}=3\)
d)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{6}+\sqrt{9}+\sqrt{12})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{2}.\sqrt{3}+\sqrt{3}.\sqrt{3}+\sqrt{3}.\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{3}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})(1-\sqrt{3})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1-\sqrt{3}\)
NC
18 tháng 7 2018
a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)
\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)
\(=3\sqrt{2}-2\sqrt{3}\)
b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)
\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)
\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)
c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)
\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=75-20=55\)
d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)
\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)
\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)
\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)
20 tháng 7 2018
\(\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\left(\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}{\left(\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{5+3\sqrt{2}-\left(5-3\sqrt{2}\right)}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{3+\sqrt{2}-\left(3-\sqrt{2}\right)}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{6\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\)
\(=\frac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\)
20 tháng 7 2018
\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\frac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\)
\(=\frac{10+2\sqrt{7}-6-2\sqrt{7}}{2\sqrt{2}}=\sqrt{2}\)
b)\(\frac{\sqrt{27}}{\sqrt{12}}+\frac{1}{2}\)
\(=\frac{\sqrt{3}.\sqrt{9}}{\sqrt{3}.\sqrt{4}}+\frac{1}{2}\)
\(=\frac{\sqrt{9}}{\sqrt{4}}+\frac{1}{2}\)
\(=\frac{3}{2}+\frac{1}{2}\)
\(\frac{4}{2}=2\)
a) \(\sqrt{45}.\sqrt{15}.\sqrt{27}\)
\(=\left(\sqrt{15}\right)^2.\left(\sqrt{3}\right)^2.\sqrt{9}\)
\(=15.3.3\)
\(=135\)