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ta co:A = 11.329-915/ (2.314)2
=11.329-(32)15/22.328
=11.329-330/22.328
=329.(11-3)/22.328
= 3.7/4
=21/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{11.3^{22}.3^7-9^{15}}{\left(3^{14}\right)^2}\)
\(=\frac{11.3^{29}-3^{30}}{3^{28}}\)
\(=\frac{3^{29}.\left(11-3\right)}{3^{28}}\)
\(=3.8=24\)
\(\frac{11.3^{22}.3^7-9^{15}}{ \left(3^{14}\right)^2}\)
= \(\frac{11.3.3^{21}.3^7-\left(3^2\right)^{15}}{\left(3^{14}\right)^2}\)
= \(\frac{33.3^{28}-3^{30}}{3^{28}}\)
= \(\frac{33.3^{28}-3^{28}.9}{3^{28}}\)
= \(\frac{3^{28}.\left(33-9\right)}{3^{28}}\)
= \(\frac{24}{1}=24\)
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\(A=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5-7.2\right)}\)\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(5-14\right)}\)
\(=\frac{2^{29}.3^{18}.\left(5.2-1\right)}{2^{28}.3^{18}.\left(-9\right)}=\frac{2.9}{-9}=-2\)
Vậy A=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
=\(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
=\(\frac{2^{29}\left(5.3^{18}-3^{18}\right)}{2^{28}\left(5.3^{18}-3^{18}\right).7}\)
=\(\frac{2}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\frac{5.\left(2^2.3^2\right)^6.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
= \(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2.2^{28}.3^{18}}\)
= \(\frac{5.2^{29}.2.3^{18}-2^{29}.3^{18}}{\left(5-14\right).2^{28}.3^{18}}\)
= \(\frac{\left(10-1\right).2^{29}.3^{18}}{\left(5-14\right).2^{28}.3^{18}}\)
= \(\frac{9.2}{-9}\)
= \(-2\)
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B = 11 . 322 . 37 - 915 / (2 . 314)2
B = 11 . 329 - (32)15 / 22 . 328
B = 11 . 329 - 330 / 22 . 328
B = 329 . (11 - 3) / 22 . 328
B = 329 . 8 / 22 . 328
B = 329 . 23 / 22 . 328
B = 3 . 2
B = 6