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10^30=(10^3)^10=1000^10
2^100=(2^10)^10=1024^10
=>10^30<2^100
vậy.....
5^40=(5^4)^10=625^10
=>5^40>620^10
\(a,\) Ta có : \(10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}\Rightarrow10^{30}< 2^{100}\)
\(b,\)Ta có : \(5^{40}=\left(5^4\right)^{10}=625^{10}\)
Vì \(625^{10}>620^{10}\Rightarrow5^{40}>620^{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta thấy A < 1
=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)
Vậy A < B
Bài 2:
Ta thấy C < 1
=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)
Vậy C < D
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103và 2100
Ta có:1030=(103)10=100010
2100=(210)10=102410
Vì 1000<1024 nên 1030<2100
5300 và 3453
Ta có:5300=(52)150=25150
3453=(33)151=27151=27.27150
Vì 25 < 27.27 nên 5300<3453
nhớ k ch mình nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^{36}\)và \(11^{24}\)
Ta có :
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vi \(125>121\)nên \(5^{36}>11^{24}\)
\(3^{100}\)và \(2^{150}\)
Ta có :
\(3^{100}=\left(3^2\right)^{50}=9^{50}\)
\(2^{150}=\left(2^3\right)^{50}=8^{50}\)
Vì \(9>8\)nên \(3^{100}>2^{150}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\) Ta có :
\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)
\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)
Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)
Do đó :
\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(10^{30}=\left(10^3\right)^{10}=1000^{10}< 1024^{10}=\left(2^{10}\right)^{10}=2^{100}\\ \)
\(2^{100}\) và \(18^{120}\)
\(2^{100}=\left(2^5\right)^{20}=32^{20}\)
\(18^{120}=\left(18^6\right)^{20}=34012224^{20}\)
Vì \(32< 34012224\)
Vậy \(2^{100}< 18^{120}\)