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áp dụng tc \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{a+m}< 1\left(m\in N\right)\)
Ta có: \(B=\frac{15^{16}+1}{15^{17}+1}< \frac{15^{16}+1+14}{15^{17}+1+14}\)\(=\frac{15^{16}+15}{15^{17}+15}=\frac{15.\left(15^{15}+1\right)}{15.\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}\)
\(\Rightarrow B< A\)
\(A=\frac{15^{15}+1}{15^{16}+1}\)
\(\Rightarrow15A=\frac{15^{16}+15}{15^{16}+1}\)
\(\Rightarrow15A=\frac{15^{16}+1+14}{15^{16}+1}\)
\(\Rightarrow15A=\frac{15^{16}+1}{15^{16}+1}+\frac{14}{15^{16}+1}\)
\(\Rightarrow15A=1+\frac{14}{15^{16}+1}\)
\(B=\frac{15^{16}+1}{15^{17}+1}\)
\(\Rightarrow15B=\frac{15^{17}+15}{15^{17}+1}\)
\(\Rightarrow15B=\frac{15^{17}+1+14}{15^{17}+1}\)
\(\Rightarrow15B=\frac{15^{17}+1}{15^{17}+1}+\frac{14}{15^{17}+1}\)
\(\Rightarrow15B=1+\frac{14}{15^{17}+1}\)
Vì \(\frac{14}{15^{17}+1}< \frac{14}{15^{16}+1}\) nên \(15B< 15A\)
Vậy B < A
Ta có công thức \(\frac{a}{b}<1\)thì\(\frac{a}{b}<\frac{a+n}{b+n}\)
\(B=\frac{15^{16}+1}{15^{17}+1}<\frac{15^{16}+1+14}{15^{17}+1+14}=\frac{15^{16}+15}{15^{17}+15}=\frac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}=A\left(1\right)\)
từ (1) \(\Leftrightarrow A>B\)
a, Vì A, B < 1
\(A=\frac{15^{16}+1}{15^{17}+1}< \frac{15^{16}+1+14}{15^{17}+1+14}=\frac{15^{16}+15}{15^{17}+15}=\frac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}\)
b, \(B=\frac{2018^{2018}+1}{2018^{2019}+1}< 1< \frac{2018^{2019}+1}{2018^{2018}+1}=A\)
A=10^15+1/10^16+1
=>10A=1+9/10^16+1
B=10^16+1/10^17+1
=>10B=1+9/10^17+1
=>10A>10B=>A>B
Vậy:A>B
a) 2317 - 2316 = 2316(23-1) = 2316.22
2316 - 2315 = 2315( 23-1) = 2315.22
Do........... ( tới đây chắc bn làm đc)
b) 1719 + 1717 = 1717( 172+ 1) = 1717. ( 289+1) = 1717 . 290
2 .1718 = 1717 . (17.2) = 1717. 34
Do..................
Vì 1517 < 1519 => 1517 + 1 < 1519 + 1 => a < b
@may la ai