Rút gọn các biểu thức sau:\(\frac{\frac{1}{2}-\frac{1}{3}+\frac{2}{5}+\frac{1}{8}}{\frac{1}{6}+\frac{3}{20}-\frac{1}{3}+\frac{3}{4}}\)
K
Khách
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18 tháng 8 2023
a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)
b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)
c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)
\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)
LH
2 tháng 1 2016
a, Dat A =\(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\)
\(\Rightarrow\frac{1}{3}A+A=\left(\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\right)+\left(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\right)\)
\(\Rightarrow\frac{4}{3}A=\frac{1}{3}+\frac{1}{3^{200}}\)
\(\Rightarrow A=\frac{\frac{1}{3}+\frac{1}{3^{200}}}{\frac{4}{3}}\)
chung minh tuong tu cau b va c
4 tháng 5 2016
Với mọi \(k\ge2\) thì \(\frac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}=\frac{\left[\left(\sqrt{k-1}\right)^2+\left(\sqrt{k+1}\right)^2+\sqrt{\left(k-1\right)\left(k+1\right)}\right]\left(\sqrt{k+1}-\sqrt{k-1}\right)}{\left(\sqrt{k-1}+\sqrt{k+1}\right)\left(\sqrt{k+1}-\sqrt{k-1}\right)}\)
\(=\frac{\sqrt{\left(k+1\right)^3}-\sqrt{\left(k-1\right)^3}}{2}\)
Suy ra tổng đã cho có thể viết là :
\(A=\frac{1}{2}\left[\sqrt{3^3}-\sqrt{1^3}+\sqrt{4^3}-\sqrt{2^3}+\sqrt{5^3}-\sqrt{3^3}+\sqrt{6^3}-\sqrt{4^3}+...+\sqrt{101^3}-\sqrt{99^3}\right]\)
\(=\frac{1}{2}\left[-1-\sqrt{2^3}+\sqrt{101^3}+\sqrt{100^3}\right]\)
\(=\frac{999+\sqrt{101^3}-\sqrt{8}}{2}\)
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
a) Cách 1:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)
Cách 2:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)
b)
\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)
8 tháng 4 2016
Xét mẫu số ta có: \(2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
=\(2012+\left(\frac{2014-2}{2}+\frac{2014-3}{3}+...+\frac{2014-2013}{2013}\right)\)
= \(2012+\left(\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+...+\frac{2014}{2013}\right)-\left(\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+...+\frac{2013}{2013}\right)\)
= \(2012+2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)-2012\)
= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)\)
\(\Rightarrow A=\frac{1}{2014}\)
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