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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\left(17^5\right)^2=17^{10}\)
Vì \(17^{10}=17^{10}\Rightarrow\left(17^5\right)^2=17^{10}\)
\(2^{60}=\left(2^3\right)^{20}=8^{20}\)
Vì \(8^{20}>4^{20}\Rightarrow2^{60}>4^{20}\)
\(5^{45}=\left(5^3\right)^{15}=125^{15}\)
Vì \(125^{15}>25^{15}\Rightarrow5^{45}>25^{15}\)
a, Ta có : \(\left(17^5\right)^2=17^{10}\)
Vì \(17^{10}=17^{10}\Rightarrow\left(17^5\right)^2=17^{10}\)
b, Ta có \(2^{60}=\left(2^3\right)^{20}=8^{20}\)
Vì \(8^{20}>4^{20}\Rightarrow2^{60}>4^{20}\)
c, Ta có : \(5^{45}=\left(5^3\right)^{15}=125^5\)
Vì \(125^5>25^{15}\Rightarrow5^{45}>25^{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
60^5 và 15^10 ta có: 15^10= (15^2)^5= 225^5
=> 60^5 >15^10
1/20^7 và 1/5^9 ta có 20^7>5^9
=> 1/20^7 <1/5^9
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=\(\left[\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}+\dfrac{2^9.2^6.3^8}{\left(3.2\right)^6.2^9}\right]=\left[\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}+\dfrac{2^6.3^8}{3^6.2^6}\right]\)
=\(\left[\dfrac{2^5}{0,4}+3^2\right]\)
=\(\left[80+9\right]=89\)
\(\left[\dfrac{\left(2.0,4\right)^5}{0,4,0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^6.2^9}\right]\div\dfrac{3^{20}.5^{30}}{3^{15}.5^{30}}\)
\(=\left[\dfrac{2^5.0.4^5}{0,4.0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^{15}}\right]\div3^5\)
\(=\left[\dfrac{2^5}{0,4}+3^2\right]\div243\)
\(=80+\left(3^5\div3^2\right)\)
\(=80+3^3\)
\(=80+27\)
\(=107\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2^{30}=8^{10}< 9^{10}=3^{20}\)
b: \(25^{15}=5^{30}< 5^{45}\)
c: \(2^{60}=\left(2^3\right)^{20}=8^{20}< 9^{20}=3^{40}\)
d: \(64^8=4^{24}\)
\(16^{12}=\left(4^2\right)^{12}=4^{24}\)
Do đó: \(64^8=16^{12}\)
e: \(2^{300}=8^{100}< 9^{100}=3^{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) 3x = 45 + 15 = 60
x = 60 : 3 = 20
2) 5x = 50 - 35 = 15
x = 15 : 5 = 3
3) (2x - 5) + 17 = 6
2x - 5 = 6 - 17
2x - 5 = -11
2x = -11 + 5
2x = -6
x = -6 : 2 = -3
4) 10 - 2(4 -3x) = -4
2(4 - 3x) = -4 - 10 = -14
4 - 3x = -14 : 2 = -7
3x = -7 - 4 = -18
x = -18 : 3 = -6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}3x-15=45\Leftrightarrow3x=60\Leftrightarrow x=20\\35-5x=50\Leftrightarrow5x=-15\Leftrightarrow x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)+17=6\Leftrightarrow2x+5=-11\Leftrightarrow2x=-16\Leftrightarrow x=-8\\10-2\left(4-3x\right)=-4\Leftrightarrow8-6x=14\Leftrightarrow6x=-6\Leftrightarrow x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}-12+3\left(-x+7\right)=-18\Leftrightarrow-3x+21=-6\Leftrightarrow-3x=-27\Leftrightarrow x=9\\24:\left(3x-2\right)=-3\Leftrightarrow3x-2=-8\Leftrightarrow3x=-6\Leftrightarrow x=-2\end{matrix}\right.\\-45:5\left(-3-2x\right)=3\Leftrightarrow-15-10x=-15\Leftrightarrow10x=0\Leftrightarrow x=0\end{matrix}\right.\)
+) Ta có: (175)2 = 175.2 = 1710
Ta thấy: 1710 = 1710 => (175)2 = 1710
+) Ta có: 420 = (22)20 = 22.20 = 240
Ta thấy: 260 > 240 => 260 > 420
+) Ta có: 2515 = (52)15 = 52.15 = 530
Ta thấy: 545 > 530 => 545 > 2515
+) Ta có: 648 = (43)8 = 43.8 = 424
1612 = (42)12 = 42.12 = 424
Ta thấy: 424 = 424 => 648 = 1612
(175)2 = 175.2 = 1710
mà 1710 = 1710
Vậy (175)2 = 1710