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\(=\frac{\left(-1\right)^7}{3^7}.3^7+\left(\frac{1}{8}\right)^3.8^3+\left(\frac{1}{5}\right)^3.\left(2.5\right)^3\)
\(=\left(-1\right)^7+\frac{1^3}{8^3}.8^3+\frac{1^3}{5^3}.2^3.5^3\)
\(=-1+1+2^3=2^3=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B1. phân a tui ko bt nha :>
\(B=\frac{2^{13}\cdot9^4}{6^6\cdot8^3}\)
\(=\frac{2^{13}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}\)
\(=\frac{2^{13}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
\(=\frac{2^{13}\cdot3^8}{2^{15}\cdot3^6}\)
\(=\frac{1\cdot3^2}{2^2\cdot1}\)
\(=\frac{1\cdot9}{4\cdot1}\)
\(=\frac{9}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
22222 - 13345 = 9877
345678 + 456456 = 801134
987888 - 111111 + 2021 = 878798
1111111 - 234456 + 234 = 879032
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a) \(-\frac{4}{15}\cdot\left(-\frac{3}{8}\right)\cdot\left(\frac{15}{-4}\right)\cdot8\)
\(=\left(-\frac{4}{15}\cdot\left(\frac{15}{-4}\right)\right)\cdot\left(-\frac{3}{8}\cdot8\right)\)
\(=1\cdot\left(-3\right)=-3\)
b) \(-\frac{2}{3}\cdot\frac{4}{19}+\frac{5}{3}\cdot\frac{4}{19}\)
\(=\frac{4}{19}\cdot\left(\frac{5}{3}-\frac{2}{3}\right)\)
\(=\frac{4}{19}\cdot1=\frac{4}{19}\)
c) \(-\frac{1}{2}\cdot\frac{5 }{3}+\frac{1}{2}\cdot\frac{8}{3}\)
\(=\frac{1}{2}\left(\frac{8}{3}-\frac{5}{3}\right)\)
\(=\frac{1}{2}\cdot1=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, bạn tự làm
b, \(B=\dfrac{5^2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)
\(=5\left(1-\dfrac{1}{106}\right)=\dfrac{5.105}{106}=\dfrac{525}{106}\)
c, đk : \(x\ne\dfrac{2}{3}\)
Ta có : \(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)(tm)
Với x = 3 suy ra \(C=\dfrac{2.9+9-1}{3.3-2}=\dfrac{26}{7}\)
Với x = -1 suy ra \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)
\(\dfrac{1}{12345}-3\dfrac{3}{4}+0,125+3\dfrac{3}{4}-\dfrac{1}{8}-\dfrac{1}{12345}\)
\(=\left(\dfrac{1}{12345}-\dfrac{1}{12345}\right)+\left(-3\dfrac{3}{4}+3\dfrac{3}{4}\right)+\left(0,125-\dfrac{1}{8}\right)\)
\(=0+0+\left(\dfrac{1}{8}-\dfrac{1}{8}\right)\)
\(=0\)
#Toru
cảm ơn bn nhiều nhé