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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 )
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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 \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)
B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)
vậy A=B
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Gọi \(\frac{13^{15}+1}{13^{16}+1}\)là S, \(\frac{13^{16}+1}{13^{17}+1}\)là X
\(13\cdot S=13\cdot\frac{13^{15+1}}{13^{16}+1}=\frac{13.\left(13^{15}+1\right)}{13^{16}+1}=\frac{13^{16}+13}{13^{16}+1}\)\(=\frac{13^{16}+1+12}{13^{16}+1}=\frac{13^{16}+1}{13^{16}+1}+\frac{12}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(13\cdot X=13.\frac{13^{16}+1}{13^{17}+1}=\frac{13\cdot\left(13^{16}+1\right)}{13^{17}+1}=\frac{13^{17}+13}{13^{17}+1}\)\(=\frac{13^{17}+1+12}{13^{17}+1}=\frac{13^{17}+1}{13^{17}+1}+\frac{12}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Do \(1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)\(\rightarrow13\cdot S>13\cdot X\)\(\rightarrow S>X\)