TA CÓ:

15/301 < 15/300 = 1/20 = 25/500 < 25/499

VÌ: 15/301 < 1/20 < 25/499 NÊN 15/301 < 25/499

25/499 > 15/301

A).Bạn lấy 15 : 301 = 0,049

B).Bạn lấy 25 : 499 = 0,050

Vậy là \(\frac{15}{301}\) <  \(\frac{25}{499}\)

\(\frac{15}{301}< \frac{25}{499}\)( dựa vào tích chéo)

\(\Rightarrow A< B\)

Ta quy đồng tử số. Ta có tử số chung là 75:

\(\frac{15}{301}=\frac{15.5}{301.5}=\frac{75}{1505}\)

\\(\frac{25}{499}=\frac{25.3}{499.3}=\frac{75}{1497}\)

Do  \(\frac{75}{1505}< \frac{75}{1497}\)

\(\Rightarrow\frac{15}{301}< \frac{25}{499}\)

\(\Leftrightarrow A< B\)

Ta quy đồng

\(\frac{15}{301}=\frac{15.499}{301.499}=\frac{7485}{150199}\)

\(\frac{25}{499}=\frac{25.301}{499.301}=\frac{7525}{150199}\)

So sánh:

\(\frac{7485}{150199}< \frac{7525}{150199}\)

\(=>\frac{15}{301}< \frac{25}{499}\)

Vậy \(\frac{15}{301}< \frac{25}{499}\)

a) Ta có:
\(\frac{15}{301}>\frac{15}{300}=\frac{1}{20}\)

\(\frac{25}{499}< \frac{25}{500}=\frac{1}{20}\)

Vì \(\frac{1}{20}=\frac{1}{20}\) nên \(\frac{15}{301}>\frac{1}{20}>\frac{25}{499}\) hay \(\frac{15}{301}=\frac{25}{499}\)

Vậy \(\frac{15}{301}>\frac{25}{499}\)

a)Ta có:\(\frac{15}{301}\)<\(\frac{15}{300}\)=\(\frac{1}{20}\)=\(\frac{25}{500}\)<\(\frac{25}{499}\)

Vì \(\frac{15}{301}\)<\(\frac{25}{500}\)<\(\frac{25}{499}\)

\(\Rightarrow\)\(\frac{15}{301}< \frac{25}{499}\)(đpcm)

Còn câu b  thì sao.

 ta có :

\(25^{1008}=\left(5^2\right)^{1008}=5^{2.1008}=5^{2016}\)

mà \(5^{2017}>5^{2016}\)

\(\Rightarrow\)\(5^{2017}>\left(5^2\right)^{1008}\)

\(\Rightarrow\)\(5^{2017}>25^{1008}\)

có \(5^{2017}=\left(5^2\right)^{1008}\times5\)\(=25^{1008}\times5\)

mà \(=25^{1008}\times5\)> \(25^{1008}\)

nên \(5^{2017}>25^{1008}\)

Ta có:

\(5^{2017}>5^{2016}=\text{[}5^2\text{]}^{1008}=25^{1008}\)

Suy ra: 5²⁰¹⁷ > 25¹⁰⁰⁸

Ta có:

\(1-A=1-\frac{10^{101}-1}{10^{102}-1}=\frac{10^{102}-1-\text{[}10^{101}-1\text{]}}{10^{102}-1}=\frac{10^{102}-1-10^{101}+1}{10^{102}-1}\)\(=\frac{10^{102}-10^{101}}{10^{102}-1}=\frac{10^{101}\left[10-1\right]}{10^{101}\text{[}10-\frac{1}{10^{101}}\text{]}}=\frac{10-1}{10-\frac{1}{10^{101}}}=\frac{9}{10-\frac{1}{10^{101}}}\)

\(1-B=1-\frac{10^{100}+1}{10^{101}+1}=\frac{10^{101}+1-\left[10^{100}+1\right]}{10^{101}+1}=\frac{10^{101}+1-10^{100}-1}{10^{100}+1}\)

\(=\frac{10^{101}-10^{100}}{10^{101}+1}=\frac{10^{100}\left[10-1\right]}{10^{100}\text{[}10+\frac{1}{10^{100}}\text{]}}=\frac{10-1}{10+\frac{1}{10^{100}}}=\frac{9}{10+\frac{1}{10^{100}}}\)

Vì \(\frac{9}{10-\frac{1}{10^{101}}}>\frac{9}{10+\frac{1}{10^{100}}}\Rightarrow A< B\)

giúp mình trả lời ngay hôm nay nha

a)  \(\left(5+8\right)^{100}=13^{100}\)

\(\left(25-12\right)^{101}=13^{101}\)

vi \(13^{100}< 13^{101}\)nen \(\left(5+8\right)^{100}< \left(25-12\right)^{101}\)

b) \(\left(15-8\right)^{10}=7^{10}\)

\(7^{11}=7^{11}\)

vi \(7^{11}>7^{10}\)nen \(\left(15-8\right)^{10}< 7^{11}\)

a, (5+8)¹⁰⁰ và (25-12)¹⁰¹

(5+8)¹⁰⁰=13¹⁰⁰

(25-12)¹⁰¹=13¹⁰¹

Vì 13¹⁰⁰ < 13¹⁰¹

=> (5+8)¹⁰⁰ < (25-12)¹⁰¹

b, (15-8)¹⁰ và 7¹¹

(15-8)¹⁰ =7¹⁰

Vì 7¹⁰ < 7¹¹

=> (15-8)¹⁰ < 7¹¹

A= abx100 + 25 + 2500 + ab x1= abx( 100 +1) + 2500 +25= abx101 + 2525= ab x101 + 25 x101 = 101 x(ab + 25)

B= (ab + 25 ) x 101

Vậy A = B