Question

[Hoàng Khánh Nhi]

So Sánh

a, \(\frac{21}{52}\)và \(\frac{213}{523}\)                                               b, \(\frac{1515}{9797}\) và \(\frac{171171}{991991}\)

c, \(\frac{n+2}{n+3}\)và \(\frac{n+3}{n+4}\) (n thuộc N khác 0)                      d, \(\frac{n+7}{n+6}\)và\(\frac{n+1}{n}\)

e, \(\frac{5}{7^3}\)+\(\frac{3}{7^4}\)và \(\frac{3}{7^3}\)+ \(\frac{5}{7^4}\)

f,  \(\frac{9^{2019}+1}{9^{2020}+1}\)và \(\frac{9^{2019}-1}{9^{2020}-1}\)     

Answer 1

[Nobi Nobita]

a) \(\frac{21}{52}=\frac{210}{520}=1-\frac{310}{520}\)

\(\frac{213}{523}=1-\frac{310}{523}\)

Vì \(520< 523\)\(\Rightarrow\frac{1}{520}>\frac{1}{523}\)\(\Rightarrow\frac{310}{520}>\frac{310}{523}\)

\(\Rightarrow1-\frac{310}{520}< 1-\frac{310}{523}\)

hay \(\frac{21}{52}< \frac{213}{523}\)

b) \(\frac{1515}{9797}=\frac{15.101}{97.101}=\frac{15}{97}\); \(\frac{171171}{991991}=\frac{171.1001}{991.1001}=\frac{171}{991}\)

Ta có: \(\frac{15}{97}=\frac{150}{970}=1-\frac{820}{970}\); \(\frac{171}{991}=1-\frac{820}{991}\)

Vì \(970< 991\)\(\Rightarrow\frac{1}{970}>\frac{1}{991}\)\(\Rightarrow\frac{820}{970}>\frac{820}{991}\)

\(\Rightarrow1-\frac{820}{970}< 1-\frac{920}{991}\)

hay \(\frac{1515}{9797}< \frac{171171}{991991}\)

c) \(\frac{n+2}{n+3}=1-\frac{1}{n+3}\); \(\frac{n+3}{n+4}=1-\frac{1}{n+4}\)

Vì \(n\inℕ^∗\)\(\Rightarrow n+3< n+4\)\(\Rightarrow\frac{1}{n+3}>\frac{1}{n+4}\)

\(\Rightarrow1-\frac{1}{n+3}< 1-\frac{1}{n+4}\)

hay \(\frac{n+2}{n+3}< \frac{n+3}{n+4}\)

Answer 2

[Nobi Nobita]

d) \(\frac{n+7}{n+6}=1+\frac{1}{n+6}\); \(\frac{n+1}{n}=1+\frac{1}{n}\)

Vì \(n\inℕ^∗\)\(\Rightarrow n+6>n\)\(\Rightarrow\frac{1}{n+6}< \frac{1}{n}\)

\(\Rightarrow1+\frac{1}{n+6}< 1+\frac{1}{n}\)

hay \(\frac{n+7}{n+6}< \frac{n+1}{n}\)