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

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

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

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

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

\(\Leftrightarrow17A< 17B\)

\(\Leftrightarrow A< B\)

Trả lời

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

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

Vì \(17^{19}+1>17^{18}+1\)

\(\Rightarrow\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)

\(\Rightarrow1+\frac{16}{17^{18}+1}>1+\frac{16}{17^{19}+1}\)

\(\Rightarrow B>A\)

bn viết thế khó hiểu lắm

viết lại đi mik giải cho

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

\(\Leftrightarrow17A=\frac{17^{19}+17}{17^{19}+1}\)

\(\Leftrightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)

\(\Leftrightarrow17A=\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}\)

\(\Leftrightarrow17A=1+\frac{16}{17^{19}+1}\)

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

\(\Leftrightarrow17B=\frac{17^{18}+17}{17^{18}+1}\)

\(\Leftrightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)

\(\Leftrightarrow17B=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}\)

\(\Leftrightarrow17B=1+\frac{16}{17^{18}+1}\)

Vì \(1+\frac{16}{17^{18}+1}>1+\frac{16}{17^{19}+1}\) nên 17B > 17A

Suy ra B > A

áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)

Ta có: \(A=\frac{17^{18}-1}{17^{20}-1}< \frac{17^{18}-1-16}{17^{20}-1-16}\)\(=\frac{17^{18}-17}{17^{20}-17}=\frac{17.\left(17^{17}-1\right)}{17.\left(17^{19}-1\right)}\)\(=\frac{17^{17}-1}{17^{19}-1}\)

\(\Rightarrow A< B\)

\(A=\frac{17^{18}-1}{17^{20}-1}\Rightarrow17^2A=\frac{17^{18}-1}{17^{18}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}\left(1\right)\)

\(B=\frac{17^{17}-1}{17^{19}-1}\Rightarrow17^2B=\frac{17^{17}-1}{17^{17}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(2\right)\)

\(\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}< \frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\Rightarrow1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}>1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(3\right)\)

Từ \(\left(1\right);\left(2\right)\&\left(3\right)\Rightarrow17^2A>17^2B\Leftrightarrow A>B.\)

\(A=\frac{17^{18}-1}{17^{20}-1}\)

\(17^2A=\frac{17^2\left(17^{18}-1\right)}{17^{20}-1}=\frac{17^{20}-17^2}{17^{20}-1}=\frac{17^{20}-1-288}{17^{20}-1}=1-\frac{288}{17^{20}-1}\)

\(B=\frac{17^{17}-1}{17^{19}-1}\)

\(17^2B=\frac{17^2\left(17^{17}-1\right)}{17^{19}-1}=\frac{17^{19}-17^2}{17^{19}-1}=\frac{17^{19}-1-288}{17^{19}-1}=1-\frac{288}{17^{19}-1}\)

Ta có : \(\frac{288}{17^{20}-1}< \frac{288}{17^{19}-1}\)nên \(-\frac{288}{17^{20}-1}>-\frac{288}{17^{19}-1}\)

\(\Rightarrow A>B\)

a>b 

tick nhé

A>B

Tick mk vài cái lên 260 nha !!!

Cơn Mưa có j để chứng minh ko

ta có A=\(\frac{17^{18}+1}{17^{19}+1}\)<\(\frac{17^{18}+1+16}{17^{19}+1+16}\) (nếu a/b<1 thì a+c/b+c>a/b)

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

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

hay A<B

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

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

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

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

Do \(17^{19}+1< 17^{19}+10\Rightarrow1-A>1-B\)

                  \(\Rightarrow A< B\)

17/18 - 1/6 bằng bao nhiêu

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

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

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

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

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

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

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

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

\(\Rightarrow17A< 17B\)

\(\Rightarrow A< B\)