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_ Dạ đừng ai tk e vỳ cái nk trên kia là của e , e đăng cho đứa pn thoy _
Bài giải
\(A=\dfrac{2017^{17}+1}{2017^{16}+1}=\dfrac{2017^{17}+2017-2016}{2017^{16}+1}=\dfrac{\left(2017.2017^{16}\right)+\left(2017.1\right)-2016}{2017^{16}+1}=\dfrac{2017.\left(2017^{16}+1\right)}{\left(2017^{16}+1\right)}=\dfrac{2017.2017^{16}+1}{2017^{16}+1}-\dfrac{2016}{2017^{16}+1}=2017-\dfrac{2016}{2017^{16}+1}\)
\(B=\dfrac{2017^{18}+1}{2017^{17}+1}=\dfrac{2017^{18}+2017-2016}{2017^{17}+1}=\dfrac{\left(2017.2017^{17}+2017.1\right)-2016}{2017^{17}+1}=\dfrac{2017.\left(2017^{17}+1\right)-2016}{2017^{17}+1}=\dfrac{2017.\left(2017^{17}+1\right)-2016}{2017^{17}+1}=\dfrac{2017.\left(2017^{17}+1\right)}{2017^{17}+1}-\dfrac{2016}{2017^{17}+1}=2017-\dfrac{2016}{2017^{17}+1}\)
Vì \(\dfrac{2016}{2017^{16}+1}>\dfrac{2016}{2017^{17}+1}\)
\(\Rightarrow2017-\dfrac{2016}{2017^{16}+1}< 2017-\dfrac{2016}{2017^{17}+1}\)
\(\Rightarrow A< B\)
Ta có:\(\frac{2017^{18}+1}{2017^{17}+1}>1\)
\(\Rightarrow\frac{2017^{18}+1}{2017^{17}+1}>\frac{2017^{18}+1+2016}{2017^{17}+1+2016}=\frac{2017^{18}+2017}{2017^{17}+2017}\)\(=\frac{2017\left(2017^{17}+1\right)}{2017\left(2017^{16}+1\right)}=\frac{2017^{17}+1}{2017^{16}+1}\)
Vậy \(\frac{2017^{17}+1}{2017^{16}+1}< \frac{2017^{18}+1}{2017^{17}+1}\)
Thanks you nhiều nha,lần sau nhớ giải hộ mình các bài toán khác nữa nha
a)\(\dfrac{17}{15}>1;\dfrac{29}{37}< 1\Leftrightarrow\dfrac{17}{15}>\dfrac{29}{37}\)
b) \(\dfrac{13}{17}>\dfrac{13}{18}\Leftrightarrow\dfrac{13}{17}>\dfrac{12}{18}\)
d)\(1-\dfrac{2017}{2018}=\dfrac{1}{2018}\)
\(1-\dfrac{2018}{2019}=\dfrac{1}{2019}\)
\(\dfrac{1}{2018}>\dfrac{1}{2019}\Leftrightarrow\dfrac{2017}{2018}< \dfrac{2018}{2019}\)
e) \(\dfrac{2018}{2017}< 1;\dfrac{2019}{2018}>1\Leftrightarrow\dfrac{2018}{2017}< \dfrac{2019}{2018}\)
Các câu dễ bạn tự làm nha:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2017^{2017}+1}{2017^{2018}+1}< 1\)
\(A< \dfrac{2017^{2017}+1+2016}{2017^{2018}+1+2016}\Rightarrow A< \dfrac{2017^{2017}+2017}{2017^{2018}+2017}\Rightarrow A< \dfrac{2017\left(2017^{2016}+1\right)}{2017\left(2017^{2017}+1\right)}\Rightarrow A< \dfrac{2017^{2016}+1}{2017^{2017}+1}=B\)\(A< B\)
Ta có :
\(2017A=\dfrac{2017\left(2017^{2015}+1\right)}{2017^{2016}+1}\)
\(=\dfrac{2017^{2016}+2017}{2017^{2016}+1}\)
\(=\dfrac{\left(2017^{2016}+1\right)+2016}{2017^{2016}+1}\)
\(=\dfrac{2017^{2016}+1}{2017^{2016}+1}\) + \(\dfrac{2016}{2017^{2016}+1}\)
\(=1+\dfrac{2016}{2017^{2016}+1}\) (1)
Tương tự :
\(2017B=\dfrac{2017\left(2017^{2014}+1\right)}{2017^{2015}+1}\)
\(=\dfrac{2017^{2015}+2017}{2017^{2015}+1}\)
\(=1+\dfrac{2016}{2017^{2016}+1}\) (2)
Từ (1) và (2) => \(2017A< 2017B\)
=> \(A< B\)
Vì \(B=\dfrac{2017^{2018}-2}{2017^{2019}-2}< 1\)
Ta có :
\(B=\dfrac{2017^{2018}-2}{2017^{2019}-2}< \dfrac{2017^{2018}-2+2019}{2017^{2019}-2+2019}=\dfrac{2017^{2018}+2017}{2017^{2019}+2017}=\dfrac{2017\left(2017^{2017}+1\right)}{2017\left(2017^{2018}+1\right)}=\dfrac{2017^{2017}+1}{2017^{2018}+1}=A\)
Vậy B < A
\(A=1\dfrac{1}{15}.1\dfrac{1}{16}.1\dfrac{1}{17}......1\dfrac{1}{2016}.1\dfrac{1}{2017}\)
\(A=\dfrac{16}{15}.\dfrac{17}{16}.\dfrac{18}{17}......\dfrac{2017}{2016}.\dfrac{2018}{2017}\)
\(A=\dfrac{16.17.18......2017.2018}{15.16.17......2016.2017}\)
\(A=\dfrac{2018}{15}\)
a) \(S=\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
\(\Rightarrow2S=\dfrac{2\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)
\(2S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}\)
\(\Rightarrow2S-S=S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}-\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
\(S=\dfrac{\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)
\(S=\dfrac{2^{2018}-2}{1-2^{2017}}=\dfrac{-2\left(1-2^{2017}\right)}{1-2^{2017}}=-2\) vậy \(S=-2\)
\(B=\dfrac{2017^{18}+1}{2017^{17}+1}< \dfrac{2017^{18}+1+2016}{2017^{17}+1+2016}\)
Mà \(\dfrac{2017^{18}+1+2016}{2017^{17}+1+2016}=\dfrac{2017^{18}+2017}{2017^{17}+2017}=\dfrac{2017.\left(2017^{17}+1\right)}{2017.\left(2017^{16}+1\right)}=\dfrac{2017^{17}+1}{2017^{16}+1}=A\)
=> B < A hay :
A < B
==' pn suy suy ra hay thật B < A hay A > B chứ -,-