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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Dễ thấy:
\(-\frac{1274}{2530}< 0< 1,2\)
\(\Rightarrow-\frac{1274}{2530}< 1,2\)
b)
Ta có :
\(25^{15}=\left(5^2\right)^{15}=5^{30}\)
\(8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}=2^{30}.3^{30}=6^{30}\)
Vì \(5^{30}< 6^{30}\)
=> \(25^{15}< 8^{10}.3^{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(25^{15}=\left(5^2\right)^{15}=25^{30}\)
\(\left(-8\right)^{10}.3^{30}=8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}=2^{30}.3^{30}=\left(2.3\right)^{30}=6^{30}\)
Mà \(25^{30}>6^{30}\) nên \(25^{15}>\left(-8\right)^{10}.3^{30}\)
Vậy \(25^{15}>\left(-8\right)^{10}.3^{30}\)
ta có 25^15=(5^2)^15=5^30
(-8)^10 . 3^30=(2^3)^10 . 3^30=2^30 . 3^30=6^30
vì 5<6 nên 5^30<6^30
vậy 25^15 > (-8)^10 . 3^30
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}=\left(\frac{2}{7}\right)^{30}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}=\left(\frac{3}{7}\right)^{30}\)
Vì \(\frac{2}{7}<\frac{3}{7}\Rightarrow\left(\frac{2}{7}\right)^{30}<\left(\frac{3}{7}\right)^{30}\)
Vậy \(\frac{4^{15}}{7^{30}}<\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 777715 = (3888,52)15
= 3888,530
mà 3888,5 > 77 . Vậy 3888,530 > 7730 hay 7730 < 777715
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4^{30}=2^{60}\)
\(8^{12}=2^{36}\)
mà 60>36
nên \(4^{30}>8^{12}\)
ta có:
25^15=(5^2)^15=5^30
8^10x3^30=(2^3)^10x3^30=2^30x3^30=(2x3)^30=6^30
vì 5>6 nên 5^30>6^30
vậy 25^15>8^10x3^30