- \(\frac{1}{9}^{40}:\frac{1}{3}^{10}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)
\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)
=>-9/10=-9/10(luôn đúng)
b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)
=>347x+780=1552
=>347x=772
hay x=772/347
![](https://rs.olm.vn/images/avt/0.png?1311)
1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)
=4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)
= 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)
=4-\(\left(1-\frac{1}{9}\right)\)
= 4-\(\frac{8}{9}\)
= \(\frac{7}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{7}{13};\frac{7}{12};\frac{7}{10};\frac{7}{8}\)
\(b,\frac{9}{40};\frac{1}{4};\frac{3}{10};\frac{3}{8}\)
\(\left(\frac{1}{9}\right)^{40}:\left(\frac{1}{3}\right)^{10}=\frac{1}{9^{40}}\cdot\frac{3^{10}}{1}=\frac{3^{10}}{\left(3^2\right)^{40}}=\frac{3^{10}}{3^{80}}=\frac{1}{3^{70}}\)