Nếu nghĩ kĩ thì thấy bài này cũng đơn giản thôi.Thử xem cách giải của mk nè:

Giải: Ta có: A=\(\frac{17^{18}+1}{17^{19}+1}\)                                                        B=\(\frac{17^{17}+1}{17^{18}+1}\)

               17A=\(\frac{17^{19}+17}{17^{19}+1}\)                                                 17B=\(\frac{17^{18}+17}{17^{18}+1}\)

             17A=\(\frac{\left(17^{19}+1\right)+16}{17^{19}+1}\)                                       17B=\(\frac{\left(17^{18}+1\right)+16}{17^{18}+1}\)

               17A=\(\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}\)                             17B=\(\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}\)

               17A=\(1+\frac{16}{17^{19}+1}\)                                            17B= \(1+\frac{16}{17^{18}+1}\)

 Lại có: 17¹⁹+1>17¹⁸+1

 Suy ra:\(\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)

             17A<17B

             A<B

Vậy A<B

\(\text{Ta có:}\frac{17^{18}+1}{17^{19}+1}\)

\(\Rightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)

\(\Rightarrow17A=1+\frac{16}{17^{19}+1}\)

\(B=\frac{17^{17}+1}{17^{18}+1}\)

\(\Rightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)

\(\Rightarrow17B=1+\frac{16}{17^{18}+1}\)

\(\text{Vì }\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)

\(\Rightarrow17A< 17B\)

\(\Rightarrow A< B\)

19 . 6 = 114; (–17) . (–10) = 17 . 10 = 170.

Vì 114 < 170 nên 19 . 6 < (–17) . (–10)

Ta có công thức :

\(\frac{a}{b}< \frac{a+c}{b+c}\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}\)

\(=\frac{17^{18}+17}{17^{19}+17}\)

\(=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}\)

\(\Leftrightarrow\frac{17^{17}+1}{17^{18}+1}\)'

\(\Rightarrow=B\)

Vậy \(A< B\)

Tham khảo :

\(17A=\dfrac{17^{19}+17}{17^{19}+1}=\dfrac{\left(17^{19}+1\right)+16}{17^{19}+1}=\dfrac{17^{19}+1}{17^{19}+1}+\dfrac{16}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)

\(17B=\dfrac{17^{18}+17}{17^{18}+1}=\dfrac{\left(17^{18}+1\right)+16}{17^{18}+1}=\dfrac{17^{18}+1}{17^{18}+1}+\dfrac{16}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)

Vì \(17^{19}>17^{18}=>17^{19}+1>17^{18}+1\)

\(=>\dfrac{16}{17^{19}+1}< \dfrac{16}{17^{18}+1}\)

\(=>17A< 17B=>A< B\)

Ta có : \(17^{17}-2< 17^{18}-2\)

Mà mẫu số càng lớn thì p/s càng bé

\(\Rightarrow\)\(\frac{2}{17^{17}-2}< \frac{2}{17^{18}-2}\)

Lại có :\(17^{18}< 17^{19}\)

\(\Rightarrow\)\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)​​​​\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)​​\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)​​\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)​​\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)( Vì số bị trừ càng lớn thì hiệu càng bé )

\(\frac{17}{19}=\frac{1717}{1919}\)

ta có : \(\frac{1717}{1919}=\frac{1717\div101}{1919\div101}=\frac{17}{19}\)

\(\Rightarrow\frac{17}{19}=\frac{1717}{1919}\)

vậy...................

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