\(\sqrt{16}\cdot\sqrt{25}+\frac{\sqrt{196}}{\sqrt{49}}\)
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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 :
\(\sqrt{3\frac{1}{16}.2\frac{14}{25}.2\frac{34}{81}}=\sqrt{\frac{49}{16}.\frac{64}{25}.\frac{196}{81}}=\sqrt{\frac{49}{16}}.\sqrt{\frac{64}{25}}.\sqrt{\frac{196}{81}}=\frac{7}{4}.\frac{8}{5}.\frac{14}{9}=\frac{196}{45}\)
a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
=4.5+14:7
=20+2
=22
b) chưa học nhó:))
a) \(\dfrac{40}{27}\)
b) \(\dfrac{196}{45}\)
c) \(\dfrac{56}{9}\)
d) 1296
a) \sqrt{\dfrac{25}{81} \cdot \dfrac{16}{49} \cdot \dfrac{196}{9}}8125⋅4916⋅9196
=\sqrt{\dfrac{25}{81}} \cdot \sqrt{\dfrac{16}{49}} \cdot \sqrt{\dfrac{196}{9}}=8125⋅4916⋅9196
=\sqrt{\left(\dfrac{5}{9}\right)^{2}} \cdot \sqrt{\left(\dfrac{4}{7}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{3}\right)^{2}}=(95)2⋅(74)2⋅(314)2
=\dfrac{5}{9} \cdot \dfrac{4}{7} \cdot \dfrac{14}{3}=\dfrac{40}{27}=95⋅74⋅314=2740.
b) \sqrt{3 \dfrac{1}{16} \cdot 2 \dfrac{14}{25} \cdot 2 \dfrac{34}{81}}3161⋅22514⋅28134
=\sqrt{\dfrac{49}{16} \cdot \dfrac{64}{25} \cdot \dfrac{196}{81}}=1649⋅2564⋅81196
=\sqrt{\dfrac{49}{16}} \cdot \sqrt{\dfrac{64}{25}} \cdot \sqrt{\dfrac{196}{81}}=1649⋅2564⋅81196
=\sqrt{\left(\dfrac{7}{4}\right)^{2}} \cdot \sqrt{\left(\dfrac{8}{5}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{9}\right)^{2}}=(47)2⋅(58)2⋅(914)2
=\dfrac{7}{4} \cdot \dfrac{8}{5} \cdot \dfrac{14}{9}=\dfrac{196}{45}=47⋅58⋅914=45196.
c) \dfrac{\sqrt{640} \cdot \sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.343}{567}}567640⋅34,3=567640.34,3=56764.343
=\sqrt{\dfrac{64.49 .7}{81.7}}=\sqrt{\dfrac{64.49}{81}}=81.764.49.7=8164.49
=\dfrac{\sqrt{64} \cdot \sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=8164⋅49=98.7
=\dfrac{56}{9}=956.
d) \sqrt{21,6} \cdot \sqrt{810} \cdot \sqrt{11^{2}-5^{2}}21,6⋅810⋅112−52
=\sqrt{21,6.810 \cdot\left(11^{2}-5^{2}\right)}=21,6.810⋅(112−52)
=\sqrt{216.81 .(11+5)(11-5)}=216.81.(11+5)(11−5)
=\sqrt{36.6 .9^{2} \cdot 4^{2} .6}=36.6.92⋅42.6
=\sqrt{36^{2} .9^{2} \cdot 4^{2}}=36.9 .4=1296=362.92⋅42=36.9.4=1296.
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
\(=20+\frac{14}{7}=20+2=22\)