Ta có: \(\dfrac{201201}{202202}=\dfrac{201}{202}\)

\(\dfrac{201201201}{202202202}=\dfrac{201}{202}\)

Do đó: \(\dfrac{201201}{202202}=\dfrac{201201201}{202202202}\)

201201/202202=201/202

201201201/202202202=201/202

Vif201/202=201/202 nên 201201/202202=201201201/202202202

Like cho me nha

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 =>\(8^{100}< 9^{100}\) =>\(3^{200}>2^{300}\)

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.

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=3^2.100=(3²)¹⁰⁰=9¹⁰⁰

2³⁰⁰=2^3.100=(2³)¹⁰⁰=8¹⁰⁰

Vì 9¹⁰⁰>8¹⁰⁰

Nên 3²⁰⁰>2³⁰⁰

b) 

Ta có: 71⁵⁰=71^2.25=(71²)²⁵=5041²⁵

37⁷⁵=37^3.25=(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

ét ô ét

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)

a. 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰

2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰

Vì 9¹⁰⁰ > 8¹⁰⁰ => 3²⁰⁰ > 2³⁰⁰

a) \(\dfrac{64}{85}\) và \(\dfrac{73}{81}\)

do \(\left\{{}\begin{matrix}64< 73\\85>81\end{matrix}\right.\)\(\Rightarrow\dfrac{64}{85}< \dfrac{73}{81}\)

b) làm tương tự bài a

c) tương tự câu a

d) \(\dfrac{25}{26}\)và\(\dfrac{25251}{26261}\)

ta có \(\dfrac{25}{26}=\dfrac{25.1010}{26.1010}=\dfrac{25250}{26260}\)

so sánh kq vừa tìm đc và 25251/26261 là đc