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a) \(\sqrt{\frac{25}{81}\cdot\frac{16}{49}\cdot\frac{169}{9}}\\ =\sqrt{\left(\frac{5}{9}\right)^2\cdot\left(\frac{4}{7}\right)^2\cdot\left(\frac{13}{3}\right)^2}\\ =\sqrt{\left(\frac{5}{9}\cdot\frac{4}{7}\cdot\frac{13}{3}\right)^2}\\ =\frac{5}{9}\cdot\frac{4}{7}\cdot\frac{13}{3}\\ =\frac{260}{189}\)
b) \(\sqrt{3\frac{1}{6}\cdot2\frac{14}{25}\cdot2\frac{34}{81}}\\ =\sqrt{\frac{19}{6}\cdot\frac{64}{25}\cdot\frac{196}{81}}\\ =\sqrt{\frac{19}{6}\cdot\left(\frac{8}{5}\right)^2\cdot\left(\frac{14}{9}\right)^2}\\ =\sqrt{\frac{19}{6}\cdot\left(\frac{8}{5}\cdot\frac{14}{9}\right)^2}\\ =\sqrt{\frac{19}{6}\cdot\frac{112}{45}}\\ =\sqrt{\frac{1064}{135}}\)
Bổ sung câu b :
\(\sqrt{3\frac{1}{16}.2\frac{14}{25}.2\frac{34}{81}}=\sqrt{\frac{49}{16}.\frac{64}{25}.\frac{196}{81}}=\sqrt{\frac{49}{16}}.\sqrt{\frac{64}{25}}.\sqrt{\frac{196}{81}}=\frac{7}{4}.\frac{8}{5}.\frac{14}{9}=\frac{196}{45}\)
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) \(-0,7\sqrt{\left(-0,7\right)^2}=-0,49\)
\(a,\sqrt{2,5}.\sqrt{30}.\sqrt{48}\)
\(=\sqrt{2,5.30.48}\)
\(=\sqrt{3600}\)
\(=60\)
\(b,\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)
\(c,\sqrt{3\frac{1}{16}.2\frac{14}{25}.2\frac{34}{81}}\)
\(=\sqrt{\frac{49}{16}.\frac{64}{25}.\frac{196}{81}}\)
\(=\sqrt{\frac{49}{16}}.\sqrt{\frac{64}{25}}.\sqrt{\frac{196}{81}}\)
\(=\frac{7}{4}.\frac{8}{5}.\frac{14}{9}\)
\(=\frac{196}{45}\)
Đúng ko
a, bạn chỉ cần lập công thức tông quát :
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cái này bạn chỉ cần trục căn thức ở mẫu chưng minh xong áp dụng vào luôn là ra
a, kq : 4/5
b,\(1-\frac{1}{\sqrt{n+1}}\)
c,d chưa nghĩ ra
ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{\left(n+1\right)n}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{\left(n+1\right)n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
nên: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}=\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)\(=1-\frac{1}{5}=\frac{4}{5}\)
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 :
\(\sqrt{3\frac{3}{16}}\) - \(\sqrt{0,36}\) = \(\sqrt{\left(\frac{7}{4}\right)^2}\) - \(\sqrt{\left(0,6^2\right)}\) = \(\frac{7}{4}-\frac{3}{5}=\frac{23}{20}\)
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) \(\sqrt{\frac{6,4}{4,9}}=\sqrt{\frac{64}{49}}=\sqrt{\left(\frac{8}{7}\right)^2}=\frac{8}{7}\)