Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(=2\sqrt{7}-8+15\sqrt{7}-12=17\sqrt{7}-20\)
b, \(=2\sqrt{2}-10\sqrt{2}+4\sqrt{2}=-4\sqrt{2}\)
c, \(=\frac{3}{8}.\frac{4}{3}-2.\frac{2}{5}=\frac{1}{2}-\frac{4}{5}=-\frac{3}{10}\)
d, \(\sqrt{\left(\sqrt{3-1}\right)^2}-\sqrt{\left(\sqrt{3-2}\right)^2}=\sqrt{3-1}-\sqrt{3-2}=\sqrt{2}-\sqrt{1}=\sqrt{2}-1\)
e, \(\sqrt{2-3}\) không tồn tại
a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))
=\(\sqrt{2006}^2-\sqrt{2005}^2\)
=2006-2005
=1
a) \(\sqrt{\frac{196}{169}}=\frac{14}{13}\)
b) \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{8}{5}\)
c) \(\sqrt{\frac{0,36}{25}}=\frac{0,6}{5}=\frac{3}{25}\)
d) \(\sqrt{\frac{6,4}{4,9}}=\sqrt{\frac{64}{49}}=\frac{8}{7}\)
a) \(\sqrt{\frac{196}{169}}=\sqrt{\left(\frac{14}{13}\right)^2}=\frac{14}{13}\)
b) \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\sqrt{\left(\frac{8}{5}\right)^2}=\frac{8}{5}\)
c) \(\sqrt{\frac{0,36}{25}}=\sqrt{\left(\frac{0,6}{5}\right)^2}=\frac{0,6}{5}=\frac{6}{50}=\frac{3}{25}\)
d) \(\sqrt{\frac{6,4}{4,9}}=\sqrt{\frac{64}{49}}=\sqrt{\left(\frac{8}{7}\right)^2}=\frac{8}{7}\)
c) \(\frac{5}{8}+\frac{13}{10}-9+25=\frac{717}{40}\)
d) \(\sqrt{0,2^2}=\left|0,2\right|=0,2\)
e) \(\sqrt{\left(-0.3\right)^2}=0,3\)
g) \(-\sqrt{\left(-1.3\right)^2}=-1,3\)
h) \(-0,7\sqrt{\left(-0,7\right)^2}=-0,49\)
a)
Ta có :
\(\sqrt{0,16}\) + \(\sqrt{\frac{4}{25}}\) = \(\sqrt{\left(0,4^2\right)}\) + \(\sqrt{\left(\frac{2}{5}\right)^2}\) = 0,4 + \(\frac{2}{5}\) = \(\frac{2}{5}+\frac{2}{5}\) = \(\frac{4}{5}\)
b)
Ta có :
\(\sqrt{3\frac{3}{16}}\) - \(\sqrt{0,36}\) = \(\sqrt{\left(\frac{7}{4}\right)^2}\) - \(\sqrt{\left(0,6^2\right)}\) = \(\frac{7}{4}-\frac{3}{5}=\frac{23}{20}\)