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`@` `\text {Ans}`
`\downarrow`
`1,`
`a)`
`3^12` và `5^8`
\(3^{12}=\left(3^3\right)^4=9^4\)
\(5^8=\left(5^2\right)^4=25^4\)
Vì `9 < 25` `=> 25^4 > 9^4`
`=> 3^12 > 5^8`
Vậy, `3^12 > 5^8`
`b)`
`(0,6)^9` và `(-0,9)^6`
\(\left(0,6\right)^9=\left(0,6^3\right)^3=\left(0,216\right)^3\)
\(\left(-0,9\right)^6=\left[\left(-0,9\right)^2\right]^3=\left(0,81\right)^3\)
Vì `0,81 > 0,216 => (0,81)^3 > (0,216)^3`
`=> (0,6)^9 < (-0,9)^6`
Vậy, `(0,6)^9<(-0,9)^6`
1.a) Có 312 = 33.4 = 274 ;
58 = 52.4 = 254
Dễ thấy 274 > 254 nên 312 > 58
b) Có \(0,6^9=\dfrac{6^9}{10^9}=\dfrac{6^{3.3}}{10^9}=\dfrac{216^3}{10^9}\)
mà \(\left(-0,9\right)^6=0,9^6=\dfrac{9^6}{10^6}=\dfrac{9^6.10^3}{10^9}=\dfrac{9^{2.3}.10^3}{10^9}=\dfrac{81^3.10^3}{10^9}=\dfrac{810^3}{10^9}\)
Dễ thấy \(\dfrac{216^3}{10^9}< \dfrac{810^3}{10^9}\Rightarrow0,6^9< \left(-0,9\right)^6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)
\(\Rightarrow\text{ }4^{100}< 2^{202}\)
b, \(3^0=1< 5^8\)
\(3^0< 5^8\)
c, \(\left(0,6\right)^0=1\)
\(\left(-0,9\right)^6=\left(0,9\right)^6\)
\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)
d,
e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)
\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)
Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,312 và 58
Ta có:312=(33)4=274
58=(52)4=254
Vì 274>254 nên 312>58
b,(0,6)9 và (0,9)6
Ta có:(0,9)6>(0,6)6 mà (0,6)6>(0,6)9
\(\Rightarrow\)(0,6)9<(0,9)6
c,52000 và 101000
Ta có:52000=(52)1000=251000>101000
\(\Rightarrow\)52000>101000
d,?????
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : \(31^5< 32^5=\left(2^5\right)^5=2^{25}< 2^{28}=\left(2^4\right)^7=16^7< 17^7\)
\(\Rightarrow31^5< 17^7\)
b) Ta có : \(8^{12}=\left(2^3\right)^{12}=2^{36}>2^{32}=\left(2^4\right)^8=16^8>12^8\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{99}\)
\(A=\frac{1-\frac{1}{99}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
a) \(31^5< 34^5=2^5.17^5=32.17^5\)
\(17^7=17^2.17^5=289.17^5\)
\(\Rightarrow31^5< 17^7\)
b) \(12^8< 16^8=\left(2^4\right)^8=2^{32}\)
\(8^{12}=\left(2^3\right)^{12}=2^{36}\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3A-A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{1}{3^{99}}\)
\(\Rightarrow2A=1-\frac{1}{3^{99}}< 1\Rightarrow A< \frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(-0,125\right)^4=0.125^4=\left(0.5^3\right)^4=0.5^{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ( 3 - 0,6) - ( 7 + 3\(\dfrac{1}{4}\) - \(\dfrac{8}{5}\)) - ( 9 - 2\(\dfrac{1}{4}\))
= 2,4 - (7 + 3,25 - 1,6) - (9 - 2,25)
= 2,4 - 7 - 3,25 + 1,6 - 9 + 2,25
= (2,4 + 1,6) - (7+ 9) - ( 3,25 - 2,25)
= 4 - 16 - 1
= - 12 - 1
= -13
b, ( - \(\dfrac{5}{8}\) + \(\dfrac{7}{6}\) - \(\dfrac{0}{8}\)) - (\(\dfrac{5}{6}\) - \(\dfrac{7}{8}\) - 1,4) + ( \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\) + \(\dfrac{12}{5}\))
= - \(\dfrac{5}{8}\) + \(\dfrac{7}{6}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{8}\) + \(\dfrac{7}{5}\) + \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\) + \(\dfrac{12}{5}\)
= (- \(\dfrac{5}{8}\) + \(\dfrac{7}{8}\)) + (\(\dfrac{7}{6}\) - \(\dfrac{5}{6}\)) + ( \(\dfrac{7}{5}\) + \(\dfrac{12}{5}\)) + \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\)
= \(\dfrac{1}{4}\) + \(\dfrac{1}{3}\) + \(\dfrac{19}{5}\) + \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + ( \(\dfrac{1}{3}\) + \(\dfrac{5}{3}\)) + \(\dfrac{19}{5}\)
= 1 + 2 + 3,8
= 6,8
3 12 và 5 8
? 4 và ?? 4
? = 27 và ?? = 25
3 12 > 5 8
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