\(2^{150}=\left(2^5\right)^{30}=32^{30}\)(1)

\(3^{90}=\left(3^3\right)^{30}=27^{30}\)(2)

Từ (1),(2)=>\(2^{150}>3^{90}\) hay \(32^{30}>27^{30}\)

\(2^{150}=\left(2^5\right)^{30}=32^{30}\)

\(3^{90}=\left(3^3\right)^{30}=27^{30}\)

mà 32>27

nên \(2^{150}>3^{90}\)

a) \(2^{135}=2^{3.45}=\left(2^3\right)^{45}=8^{45}\)
\(3^{90}=3^{2.45}=\left(3^2\right)^{45}=9^{45}\)
Vì \(8^{45}< 9^{45}\)nên \(2^{135}< 3^{90}\)

b) \(4^{75}=4^{3.25}=\left(4^3\right)^{25}=64^{25}\)
\(3^{100}=3^{4.25}=\left(3^4\right)^{25}=81^{25}\)
Vì \(64^{25}< 81^{25}\)nên \(4^{75}< 3^{100}\)

c) \(4^{100}=4^{4.25}=\left(4^4\right)^{25}=256^{25}\)
\(9^{75}=9^{3.25}=\left(9^3\right)^{25}=729^{25}\)
Vì \(256^{25}< 729^{25}\)nên \(^{4^{100}< 9^{75}}\)

Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)

Vì \(9801< 9999\)nên \(9801^{10}=9999^{10}\)

Vậy \(99^{20}< 9999^{10}\)

a, 2^27 = 2^3.9 = 8^9 

3^18 = 3^2.9 = 9^9 

vì 8<9 => 8^9 < 9^9 => 2^27 < 3^18

a) 

11*3¹¹ = 1948617

3¹³ = 1594323

vì 1948617 > 1594323 nên 11*3¹¹ > 3¹³

b) 89² = 7921

88*90 = 7920

vì 7921 > 7920 nên 89² > 88*90

mình cảm ơn :3

Ta có:
\(2^{135}=\left(2^3\right)^{45}=8^{45}\)

\(3^{90}=\left(3^2\right)^{45}=9^{45}\)

Vì \(8^{45}< 9^{45}\) nên \(2^{135}< 3^{90}\)

Vậy \(2^{135}< 3^{90}\)

\(2^{135}\) và \(3^{90}\)

\(2^{135}=2^{3.45}=\left(\left(2^3\right)^{45}\right)=8^{45}\)

\(3^{90}=3^{2.45}=\left(\left(3^2\right)^{45}\right)=9^{45}\)

Vì \(8< 9\) nên \(8^{45}< 9^{45}\) hay \(2^{135}\)\(< 3^{90}\)

a) 3^40= 3^4.10=(3^4)10=81^10
11^21> 11^20=11^2.10=(11^2)10=121^10
→ 3^40< 11^21

b) 2^195=2^15.13=(2^15)13=32768^13
3^130=3^10.13= (3^10)13=59049^13
→2^195<3^130

c) 2^90=2^5.18=(2^5)18= 32^18
5^36=5^2.18=(5^2)18=25^18
→2^90>5^36

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A = \(\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{91}{90}+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+...+\left(1+\frac{1}{90}\right)\)

\(=\left(1+\frac{1}{1.2}\right)+\left(1+\frac{1}{2.3}\right)+\left(1+\frac{1}{3.4}\right)+...+\left(1+\frac{1}{9.10}\right)\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)+\left(1+1+1+...+1\right)\)(9 số hạng 1)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)+1.9\)

\(=\left(1-\frac{1}{10}\right)+9=10-\frac{1}{10}=\frac{99}{10}>\frac{98}{11}\)

a=1+1/1.2+1+1+1/2.3+....+1+1/9.10

a=1+1+...+1(9 chữ số 1)+1/1-1/2+1/2-1/3+...+1/9-1/10

a=9+1-1/10

a=9+9/10=9+0.9=9.9

b=98/11<98/10=9.8<9.9

=>vậy a>b