Ta có: \(13A=1+\frac{12}{13^{16}+1};13B=1+\frac{12}{13^{17}+1}\)

Do \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\). Nên \(13A>13B\) 

Vậy \(A>B\)

Ta có : 

\(13A=\frac{13^{16}+13}{13^{16}+1}=\frac{13^{16}+1+12}{13^{16}+1}=\frac{13^{16}+1}{13^{16}+1}+\frac{12}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)

\(13B=\frac{13^{17}+13}{13^{17}+1}=\frac{13^{17}+1+12}{13^{17}+1}=\frac{13^{17}+1}{13^{17}+1}+\frac{12}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)

Vì \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\) nên \(1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\) hay \(13A>13B\)

\(\Rightarrow\)\(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~ 

Phùng Minh Quân ơi tớ cảm ơn nhưng tớ tính máy tính ra A = B ạ ( ko có ý gì đâu )

Bài 1:

Ta có:

\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)

\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)

Lại có:

\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)

\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)

Bài 2:

Ta có:

\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)

\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)

Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)

\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)

\(\Rightarrow13A>13B\Rightarrow A>B\)

      TA CÓ   CÔNG THỨC \(\frac{a}{b}