rut gon
can (12-6can3) - can (21-12can3)
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N
0
PT
2
MT
8 tháng 7 2015
\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}\)
=\(\left(\sqrt{4}\sqrt{7}-\sqrt{7}-\sqrt{12}\right).3+2\sqrt{21}\)
=\(\left(2\sqrt{7}-\sqrt{7}-\sqrt{4}\sqrt{3}\right).3+2\sqrt{21}\)
=\(\left(\sqrt{7}-2\sqrt{3}\right).3+2\sqrt{21}\)
=\(3\sqrt{7}-6\sqrt{3}+2\sqrt{21}\)
đề có sai ko nhưng kết quả ra thế
8 tháng 7 2015
\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right).3+2\sqrt{21}=\left(\sqrt{7}-2\sqrt{3}\right).3+2\sqrt{21}=3\sqrt{7}-6\sqrt{3}+2\sqrt{21}\)
TM
1
Q
15 tháng 3 2017
Rút gọn
A, \(\dfrac{125}{1000}\)
\(\dfrac{125}{1000}=\dfrac{125:125}{1000:125}=\dfrac{1}{8}\)
B, \(\dfrac{3600-75}{4800-175}\)
\(\dfrac{3600-75}{4800-175}=\dfrac{3525}{4625}\)
\(\Rightarrow\dfrac{3525}{4625}=\dfrac{3525:25}{4625:25}=\dfrac{141}{185}\)
C, \(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}\)
\(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\dfrac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\dfrac{\left(-5\right)}{3^2}=\dfrac{-5}{9}\)
Q
4 tháng 6 2017
a) \(\sqrt{15+2\sqrt{5}-\sqrt{21-4\sqrt{5}}}\)
\(=\sqrt{15+2\sqrt{5}-\sqrt{\left(1-2\sqrt{5}\right)^2}}\)
\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)
\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)
\(=\sqrt{15+2\sqrt{5}-2\sqrt{5}+1}\)
\(=\sqrt{16}\)
\(=4\)
b) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt[4]{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt[4]{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt[4]{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt[4]{5-\sqrt{\left(1-\sqrt{5}\right)^2}}\)
\(=\sqrt[4]{5-\left(\sqrt{5}-1\right)}\)
\(=\sqrt[4]{5-\sqrt{5}+1}\)
\(=\sqrt[4]{6-\sqrt{5}}\)
NT
9 tháng 7 2016
\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)=\(\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{3}+\sqrt{5}\right)}\)=\(\frac{\sqrt{2}}{\sqrt{7}}\)
Viết lại đề cho bạn nè
\(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
= \(\sqrt{9-6\sqrt{3}+3}-\sqrt{12-12\sqrt{3}+9}\)
= \(\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}\right)^2-2.3.\sqrt{12}+3^2}\)
= \(\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}-3\right)^2}\)
= |\(3-\sqrt{3}\)| - |\(\sqrt{12}-3\)|
= \(3-\sqrt{3}-\sqrt{12}+3\)
= \(6-\sqrt{3}-2\sqrt{3}\)
= \(6-3\sqrt{3}\)