\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}\)  theo công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)

\(A< \frac{10^{11}+10}{10^{12}+10}=\frac{10^{10}\left(10+1\right)}{10^{11}\left(10+1\right)}=\frac{10^{10}}{10^{11}}\)

\(\Rightarrow\frac{10^{10}}{10^{11}}=\frac{10^{10}\cdot10^{12}}{10^{11}\cdot10^{12}}=\frac{10^{22}}{10^{23}}\)

\(\Leftrightarrow A< \frac{10^{10}}{10^{11}}=\frac{10^{11}}{10^{12}}\)

Lại áp dụng công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)

\(A< \frac{10^{10}}{10^{11}}=\frac{10^{11}}{10^{12}}< \frac{10^{11}+1}{10^{12}+1}=B\)

\(\Leftrightarrow A< B\)

Hoặc \(A< \frac{10^{11}-1+2}{10^{12}-1+2}=\frac{10^{12}+1}{10^{12}+1}\)

..... (EZ)

\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10.\left(10^{10}+1\right)}{10.\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}\)

=> \(A< B\)

\(A=\frac{10^{11}-1}{10^{12}-1}\)

\(\Leftrightarrow10A=\frac{10\left(10^{11}-1\right)}{\left(10^{12}-1\right)}=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}\left(1\right)\)

\(B=\frac{10^{10}+1}{10^{11}+1}\)

\(\Leftrightarrow10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=\frac{9}{10^{11}+1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)

Nếu có 1  phân số a/b < 1 thì a/b < a+n/b+n.

Tương tự ta có: A < (10^11  -1)+11/(10^12 -1)+10                        

                           A < 10^11+10/10^12+10                        

                          A < 10(10^10+1)/10(10^11+1)                         

                          A < 10(10^10+1)/10(10^11+1)                        

                          A < 10^10+1/10^11+1          

                Vậy  A < B

ta có: \(A=\frac{10^{11}-1}{10^{12}-1}\Rightarrow10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}.\)

\(B=\frac{10^{11}+1}{10^{12}+1}\Rightarrow10B=\frac{10^{12}+10}{10^{12}+1}=\frac{10^{12}+1+9}{10^{12}+1}=1+\frac{9}{10^{12}+1}\)

\(\Rightarrow1-\frac{9}{10^{12}-1}< 1+\frac{9}{10^{12}+1}\Rightarrow10A< 10B\Rightarrow A< B\)

10A=10¹²-10/10¹²-1

=1-9/10¹²-1

10B=10¹²+10/10¹²+1

=1+9/10¹²+1

suy ra A<B

Ta có :

\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)

\(\Rightarrow A< B\)

bài này ko cần cách làm tớ chỉ ra kết quả thui

Ta luôn có nếu a>0; b>0 thì \(\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)

Áp dụng vào bài toán ta thấy 10¹¹-1 > 0 và 10¹²-1 > 0 nên

\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10.\left(10^{10}+1\right)}{10.\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)

 Vậy A < B

Xin lỗi bn nhé bài toán phụ phía trên đang còn 1 đk nữa là a<b

\(10A=\frac{10^{12}-10}{10^{12}-1}\); \(10B=\frac{10^{11}+10}{10^{11}+1}\)

=> \(10A=\frac{\left(10^{12}-1\right)-9}{10^{12}-1}\); \(10B=\frac{\left(10^{11}+1\right)+9}{10^{11}+1}\)

=> \(10A=\frac{10^{12}-1}{10^{12}-1}-\frac{9}{10^{12}-1}\); \(10B=\frac{10^{11}+1}{10^{11}+1}+\frac{9}{10^{11}+1}\)

=> \(10A=1-\frac{9}{10^{12}-1}\); \(10B=1+\frac{9}{10^{11}+1}\)

Ta có: \(10A=1-\frac{9}{10^{12}-1}< 1\); \(10B=1+\frac{9}{10^{11}+1}>1\)

=> \(10A< 10B\)

=> \(A< B\)

A>b nha!

để so sánh A và B ta so sánh 

 \(\frac{10^{11}-1}{10^{12}-1}\)và \(\frac{10^{10}+1}{10^{11}+1}\)

Ta có \(10^{11}-1< 10^{11}+1\)

    và  \(10^{12}-1>10^{11}+1\)

=> A<B

 \(10A=\frac{10^{12}-1-9}{10^{12}-1}=\frac{10^{12}-9}{10^{12}}-1\)

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

ta có: \(A=\frac{10^{11}-1}{10^{12}-1}\)

\(\Rightarrow10.A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=\frac{10^{12}-1}{10^{12}-1}-\frac{9}{10^{12}-1}\)\(=1-\frac{9}{10^{12}-1}< 1\)

ta có: \(B=\frac{10^{10}+1}{10^{11}+1}\)

\(\Rightarrow10.B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=\frac{10^{11}+1}{10^{11}+1}+\frac{9}{10^{11}+1}\)\(=1+\frac{9}{10^{11}+1}>1\)

\(\Rightarrow10.A< 10.B\)

\(\Rightarrow A< B\)

Ta có:

\(A=\frac{10^{11}-1}{10^{12}-1}\)

\(\Rightarrow10.A=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)

B= 10¹⁰+1/10¹¹+1

\(\Rightarrow10.B=\frac{10^{11}+10}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)

VÌ 10.B>1 và 10.A < 1

=> 10.B> 10.A

=>B>A

Vậy A<B

Có : 10A = 10.(10^11-1)/10^12-1 = 10^12-10/10^12-1 

Vì : 0 < 10^12-10 < 10^12-1 => 10A < 1 (1)

10B = 10.(10^10+1)/10^11+1 = 10^11+10/10^11+1

Vì : 10^11+10 > 10^11+1 > 0 => 10B > 1 (2)

Từ (1) và (2) => 10A < 10B

=> A < B

Tk mk nha

\(A=\frac{10^{11}-1}{10^{12}-1}\)

\(B=\frac{10^{10}+1}{10^{11}+1}\)

Mà \(\frac{10^{11}-1}{10^{12}-1}< 1\); \(\frac{10^{10}+1}{10^{11}+1}< 1\)

\(\Rightarrow\)\(A,B< 1\)

Ta có:

\(10^{11}-1>10^{10}+1\); \(10^{12}-1>10^{11}+1\)

\(\Rightarrow A>B\)

Vậy A > B

Có : 10A = 10^12-10/10^12-1 = 1 - 9/10^12-1 < 1

10B = 10^11+10/10^11+1 = 1 + 9/10^11+1 > 1

=> 10A < 10B

=> A < B

Tk mk nha

A=10¹¹-1/10¹²-1<1

=>A=10¹¹-1/10¹²-1<10¹¹+1.10/10¹²⁺1.10

                             =10¹¹+10/10¹²+10

                             =10(10¹⁰+1)/10(10¹¹+1)

                             =10¹⁰+1/10¹¹+1=B

=>A<B