Question

so sánh A=: \(\frac{17^{18}-1}{17^{20}-1}\)Và B= \(\frac{17^{17}-1}{17^{19}-1}\)

Answer 1

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

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

\(\Rightarrow A< B\)

Answer 2

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

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

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

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

Answer 3

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

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

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

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

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

\(\Rightarrow A>B\)

Answer 4

NLA Saver sai rồi

Answer 5

* Xét : \(17^2A=\frac{17^2\left(17^{18}-1\right)}{17^{20}-1}=\frac{17^{20}-17^2}{17^{20}-1}=\frac{\left(17^{20}-1\right)+\left[\left(-17\right)^2+1\right]}{17^{20}-1}\)

                        \(=\frac{17^{20}-1}{17^{20}-1}+\frac{-17^{20}+1}{17^{20}-1}=1+\frac{-17^2+1}{17^{20}-1}=1+\frac{-288}{17^{20}-1}\)

* Tương tự xét \(17^2B=1+\frac{-288}{17^{19}-1}\)

Có : 17²⁰ - 1 > 17¹⁹ - 1 => \(\frac{-288}{17^{20}-1}>\frac{-288}{17^{19}-1}\Rightarrow1+\frac{-288}{17^{20}-1}>1+\frac{-288}{17^{19}-1}\)

\(\hept{\begin{cases}17^2A>17^2B\\\text{mà}17^2>0\end{cases}}\Rightarrow A>B\)

Answer 6

ukm thanks bạn

Answer 7

Ta có:\(17^{18}-1< 17^{18}-1\)

  =>\(\frac{17^{18}-1}{17^{20}-1}< \frac{17^{18}-1-16}{17^{20}-1-16}=\frac{17^{18}-17}{17^{20}-17}=\frac{17\left(17^{17}-1\right)}{17\left(17^{19}-1\right)}=\frac{17^{17}-1}{17^{19}-1}=B\)

  =>A<B

Vậy A<B 

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nhớ k nha! thank!

   =>

Answer 8

Ta có

\(\frac{a}{b}< 1=>\frac{a}{b}>\frac{a-n}{b-n}\)

\(\frac{17^{18}-1}{17^{20}-1}>\frac{17^{18}-17}{17^{20}-17}=\frac{17\left(17^{17}-1\right)}{17\left(17^{19}-1\right)}=B\)B

=>A>B

vậy ,........

hok tốt