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a, |-2,5 + x| = 1,3
=> -2,5 + x = 1,3 hoặc -2,5 + x = -1,3
=> x = 3,8 hoặc x = 1,2
vậy_
b,
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\(a,=\dfrac{1}{6}-\dfrac{5}{6}=-\dfrac{4}{6}=-\dfrac{2}{3}\\ b,=\dfrac{3^2}{3^{12}\cdot3^{18}}=\dfrac{1}{3^{28}}\)
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- \(\dfrac{4}{27}\) - \(\dfrac{7}{12}\) - \(\dfrac{23}{27}\) + \(\dfrac{17}{12}\)
= - (\(\dfrac{4}{27}\) + \(\dfrac{23}{27}\)) + (\(\dfrac{17}{12}\) - \(\dfrac{7}{12}\))
= - \(\dfrac{27}{27}\) + \(\dfrac{10}{12}\)
= -1 + \(\dfrac{5}{6}\)
= - \(\dfrac{6}{6}\) + \(\dfrac{5}{6}\)
= - \(\dfrac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)8^{15}xx4^{13}`
`=2^{45}*2^{26}`
`=2^{71}`
`b)9^{12}xx27^{10}`
`=3^{24}xx3^{30]`
`=3^{54}`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^9}{2^9.3^9.2^{10}+\left(3.2^2\right)^{10}}\)
\(=\frac{2^{19}.3^9+5.2^{18}.3^{19}}{2^{19}.3^9+3^{10}.2^{20}}=\frac{2^{18}.3^9\left(2+5.3\right)}{2^{18}.3^9\left(2+3.2^2\right)}=\frac{17}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B = 2^27 : 4^12 = 2^27 : (2^2)^12
= 2^27 : 2^(2x12)
=2^27 : 2^24
=2^3
=8