a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)

ta có :

 \(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)

 \(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)

\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)

Vậy \(A< 3\)

a. Ta có :

\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)

\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)

\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)

Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)

Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)

Vậy \(A< 3\)

b) \(A=\frac{1+5+5^2+5^3+...+5^{10}+5^{11}}{1+5+5^2+5^3+...+5^9+5^{10}}=5^{11}\)

bn rút gọn là dc  

\(B=\frac{1+7+7^2+7^3+...+7^{10}+7^{11}}{1+7+7^2+7^3+...+7^9+7^{10}}=7^{11}\)

\(A=5^{11},B=7^{11}\)

\(\Rightarrow7^{11}>5^{11}\Rightarrow B>A\)

hk tốt #

a) Đặt A = \(\frac{5^{12}+1}{5^{13}+1}\Rightarrow5A=\frac{5^{13}+5}{5^{13}+1}=1+\frac{4}{5^{13}+1}\)

Đặt \(B=\frac{5^{11}+1}{5^{12}+1}\Rightarrow5B=\frac{5^{12}+5}{5^{12}+1}=1+\frac{4}{5^{12}+1}\)

Vì \(\frac{4}{5^{13}+1}< \frac{4}{5^{12}+1}\Rightarrow1+\frac{4}{5^{13}+1}< 1+\frac{4}{5^{12}+1}\Rightarrow5A< 5B\Rightarrow A< B\)

Áp dụng công thức : \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(a;b;m\in N\right)\)

Ta có : \(A=\frac{5^{12}+1}{5^{13}+1}< 1\)

\(\Leftrightarrow A=\frac{5^{12}+1}{5^{13}+1}< \frac{5^{12}+1+4}{5^{13}+1+4}=\frac{5^{12}+5}{5^{13}+5}=\frac{5\left(5^{11}+1\right)}{5\left(5^{12}+1\right)}=B\)

\(\Leftrightarrow A< B\)

b ) Đặt \(A=\frac{7^{10}}{1+7+7^2+...+7^9}\)   nên \(\frac{1}{A}=\frac{1+7+7^2+...+7^9}{7^{10}}\)

\(=\frac{1}{7^{10}}+\frac{7}{7^{10}}+\frac{7^2}{7^{10}}+...+\frac{7^9}{7^{10}}=\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7}\)

Đặt \(B=\frac{5^{10}}{1+5+5^2+...+5^9}\)nên \(\frac{1}{B}=\frac{1+5+5^2+...+5^9}{5^{10}}\)

\(=\frac{1}{5^{10}}+\frac{5}{5^{10}}+\frac{5^2}{5^{10}}+...+\frac{5^9}{5^{10}}=\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}\)

Ta thấy : \(\frac{1}{7^{10}}< \frac{1}{5^{10}};\frac{1}{7^9}< \frac{1}{5^9};...;\frac{1}{7}< \frac{1}{5}\)nên \(\frac{1}{A}< \frac{1}{B}\)

Vậy \(A< B\)

2/

A=1+2+2^2+...+2^10

2.A= 2+2^2+...+2^11

=>2A-A = 2^11-1=> A = 2^11 -1=B

Vậy A=B

1)5²⁰⁰³+5²⁰⁰²+5²⁰⁰¹=5²⁰⁰¹(5²+5+1)=5²⁰⁰¹(25+5+1)=5²⁰⁰¹.31

Vì 31 chia hết cho 31nên

5²⁰⁰¹.31chia hết cho 31 hay 5²⁰⁰³+5²⁰⁰²+5²⁰⁰¹ chia hết cho 31

2) A = 1+2+2²+......+2⁹+2¹⁰

=>2A=2+2²+2³+...+2¹¹

=>2A-A=2+2²+2³+...+2¹¹-(1+2+2²+...+2⁹+2¹⁰)

=>A=2¹¹-1

Vậy A=B=2¹¹-1

a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

 2 hoặc 42

Giải như mà mình không chắc nha:

a) \(A=\frac{10^8+1}{10^9+1}\)và \(\frac{10^9+1}{10^{10}+1}\)

Ta có:

  \(\frac{10^8+1}{10^9+1}\Leftrightarrow\frac{10^8+1}{10^8+10+1}\Leftrightarrow\frac{1}{10+1}=\frac{1}{11}\)

\(\frac{10^9+1}{10^{10}+1}=\frac{10^8+10+1}{10^8+10+10+1}=\frac{10+1}{10+10+1}=\frac{11}{21}\)

Ta có: \(\frac{1}{11}< \frac{11}{21}\) Vậy ......

b) Bạn giải tương tự nha! Lười lắm :v

a, Mk làm mẫu 1 bài nha !

Có : 

10A = 10^9+10/10^9+1 = 1 + 9/10^9+1

10B = 10^10+10/10^10+1 = 1 + 9/10^10+1

Vì : 10^9+1 < 10^10+1 => 9/10^9+1 > 9/10^10+1

=> 10A < 10B => A < B

Tk mk nha

Ta thấy:\(\frac{5^{11}+1}{5^{10}+1}\)>1 nên theo quy tắc : \(\frac{a}{m}\)>1 thì \(\frac{a}{m}\)>\(\frac{a+m}{b+m}\) ta có:
B=\(\frac{5^{11}+1}{5^{10}+1}\)>\(\frac{5^{11}+1+4}{5^{10}+1+4}\)>\(\frac{5^{11}+5}{5^{10}+5}\)=\(\frac{5\left(5^{10}+1\right)}{5\left(5^9+1\right)}\)=A
Vậy B>A
Nếu có gì thì cứ hỏi 
Chúc bạn học tốt!