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Mình chỉ nói sơ cái tử thôi nha!Bạn tách 6^3 thành 6^2.3;còn 3^3 tách thành 3.3^2. Như vậy bạn đặt thừa số chung ra ngoài đc rồi đó.
\(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.3^2.2^2+3^3.1}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}\)=\(\frac{3^3.13}{-13}=3^3=-27\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
Bài giải
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
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a) \(=\frac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)
b) \(=-\frac{3^3.2^3+3.3^2.2^2+3^3}{13}=-\frac{3^3.2^3+3^3.2^2+3^3}{13}=-\frac{3^3.\left(2^3+2^2+1\right)}{13}=-\frac{3^3.13}{13}=-3^3=-27\)
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\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,6\right)^5}{\left(0,2\right)^5.0,2}=\left(\frac{0,6}{0,2}\right)^5.\frac{1}{\frac{1}{5}}=3^5.5=243.5=1215\)
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3.3^5}{2^7.2^4.3^5}=\frac{3}{16}\)
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\(a.=\frac{3}{15}+\frac{-10}{15}\)
\(=-\frac{7}{15}\)
\(b.=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
\(c.=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
\(=-1+1-\frac{1}{2}\)
\(=0-\frac{1}{2}\)
\(=-\frac{1}{2}\)
\(d.=\frac{5}{6}.\left(18\frac{2}{3}-6\frac{2}{3}\right)\)
\(=\frac{5}{6}.12\)
\(=10\)
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\(\frac{6^3+3\cdot6^2+3^3}{-13}\)
\(=\frac{2^3\cdot3^3+3\cdot3^2\cdot2^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{27\cdot13}{-13}\)
\(=-27\)