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a,3^200 và 2^300
3^200=(3^2)^100=9^100
2^300=(2^3)^100=8^100
Vì 9^100>8^100=>3^200>2^300
Vậy 3^200>2^300
b, 71^50 và 37^75
71^50=(71^2)^25=5041^25
37^75=(37^3)^25=50653^25
Vì 5041^25<50653^25=> 71^50<37^75
Vậy 71^50<37^75
c, 201201/202202 và 201201201/202202202
201201201/202202202=201201/202202
=> 201201/202202=201201201/202202202
Vậy 201201/202202=201201201/202202202
a)
Ta có:3200=32.100=(32)100=9100
2300=23.100=(23)100=8100
Vì 9100>8100
Nên 3200>2300
b)
Ta có: 7150=712.25=(712)25=504125
3775=373.25=(373)25=5065325
Vì 504125<5065325
Nên 7150<3775
c)
Ta có:
201201/202202=201.1001/202.1001=201/202
201201201/202202202=201.1001001/202.1001001001= 201/202
Vì 201/202=201/202
Nên 201201/202202=201201201/202202202
a. 3200 = (32)100 = 9100
2300 = (23)100 = 8100
Vì 9100 > 8100 => 3200 > 2300
a) 3200= (32)100=9100 và 2^300=(2^3)^100=8^100
Mà 9^100>8^100
=> 3^200>2^300
b) 71^50 = (71^2)^25= 142^25 và 37^75 = (37^3)^25= 50653^25
Mà 142^25 <50653^25
=> 71^50<37^75
c) 201201/202202=201.1001/ 202.1001 = 201/202
201201201/202202202= 201. 1001001/202.1001001=201/202
=> 201201/202202=201201201/202202202.
201201/202202=201/202
201201201/202202202=201/202
Vif201/202=201/202 nên 201201/202202=201201201/202202202
Like cho me nha
Ta có:\(\dfrac{201201}{202202}\)=\(\dfrac{201}{202}\);\(\dfrac{201201201}{202202202}\)=\(\dfrac{201}{202}\)
=>\(\dfrac{201}{202}\)=\(\dfrac{201}{202}\)
=> \(\dfrac{201201}{202202}\)=\(\dfrac{201201201}{202202202}\)
Vậy: \(\dfrac{201201}{202202}\)=\(\dfrac{201201201}{202202202}\)
(quá dễ)
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
\(x-2018=\frac{201201201}{202202202}-\frac{201201}{202202}\)
\(\Rightarrow x-2018=\frac{201}{202}-\frac{201}{202}\)
\(\Rightarrow x-2018=\frac{0}{202}\)
\(\Rightarrow x-2018=0\)
\(\Rightarrow x=0+2018\)
\(\Rightarrow x=2018\)
a)\(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{21}.\frac{8}{15}\)
=\(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{7}{13}+\frac{7}{13}.\frac{8}{15}\)
=\(\frac{7}{13}.\left(\frac{7}{15}-\frac{5}{12}-\frac{8}{15}\right)\)
=\(\frac{7}{13}.\frac{7}{12}\)
=\(\frac{49}{156}\)
b)\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
=\(a.\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
=a . 0
=0
Bài 2
a)Có
\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
Vì 8<9 =>\(8^{100}< 9^{100}\) =>\(3^{200}>2^{300}\)
Bạn ấn vào câu hỏi tương tự nhé !!!! >:
Sorry