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\(B=\dfrac{2017^{18}+1}{2017^{17}+1}< \dfrac{2017^{18}+1+2016}{2017^{17}+1+2016}\)
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
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
Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
Vậy ......................
~ Học tốt ~
Ta có : \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)
\(=3+\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)
Vậy \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
Câu 1:
a, \(\left|-5\right|=5\)
b, \(\left|10\right|=10\)
c, \(\left|-5\right|-\left|10\right|=5-10=-5\)
d, -15.30= -450
Câu 2:
a, Ta có: \(\dfrac{10}{21}.\dfrac{14}{25}=\dfrac{10.14}{21.25}=\dfrac{5.2.7.2}{3.7.5.5}=\dfrac{2.2}{3.5}=\dfrac{4}{15}\)
c, Ta có: \(-\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{-5.2+3.3}{12}=\dfrac{-10+9}{12}=\dfrac{-1}{12}\)
d, \(\dfrac{11}{17}.\dfrac{3}{2017}+\dfrac{11}{17}.\dfrac{2014}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}\left(\dfrac{3}{2017}+\dfrac{2014}{2017}\right)-1\dfrac{11}{17}\)
\(=\dfrac{11}{17}.\dfrac{2017}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}-1-\dfrac{11}{17}=-1\)
Câu 7: a, Để A có nghĩa khi \(x+2\ne0\) \(\Leftrightarrow x=-2\)
b, Ta có: \(A=2\)
<=> \(\dfrac{x-1}{x+2}=2\)
<=> \(\dfrac{x-1}{x+2}-2=0\)
<=> \(\dfrac{x-1}{x+2}-\dfrac{2x+4}{x+2}=0\)
<=> \(\dfrac{x-1-2x-4}{x+2}=0\)
<=> \(\dfrac{-x-5}{x+2}=0\)
<=> -x-5=0
<=> -x=5
<=> x= -5
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\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+\dfrac{2017}{4}+...+\dfrac{2017}{2018}}{\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}}\)
Đặt \(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\) là B
\(B=\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\\ =\dfrac{2017}{1}+1+\dfrac{2016}{2}+1+...+\dfrac{1}{2017}+1-2017\\ =\dfrac{2018}{1}+\dfrac{2018}{2}+...+\dfrac{2018}{2017}-2017\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\left(2018-2017\right)\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+1\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\\ =2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2018}}{2018\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}{2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017}{2018}\)
A = \(\dfrac{2017^{2017}+1}{2017^{2018}+1}\)< \(\dfrac{2017^{2017}+2017}{2017^{2018}+2017}\)= \(\dfrac{2017.\left(2017^{2016}+1\right)}{2017.\left(2017^{2017}+1\right)}\) = \(\dfrac{2017^{2016}+1}{2017^{2017}+1}\)= B
Vậy A < B
Giải
Ta có: A =\(\dfrac{2017^{2017}+1}{2017^{2018}+1}< 1\)
\(\Rightarrow\) A< \(\dfrac{2017^{2017}+1+2016}{2017^{2018}+1+2016}\)=\(\dfrac{2017^{2017}+2017}{2017^{2018}+2017}=\dfrac{2017\left(2017^{2016}+1\right)}{2017\left(2017^{2017}+1\right)}=\dfrac{2017^{2016}+1}{2017^{2017}+1}=B\)Vậy A<B
_ 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 \(\dfrac{2016}{2017^{16}+1}>\dfrac{2016}{2017^{17}+1}\)
\(\Rightarrow2017-\dfrac{2016}{2017^{16}+1}< 2017-\dfrac{2016}{2017^{17}+1}\)
\(\Rightarrow A< B\)
cần fai mời ms lm ák , thánh ghê thật -,-