Ta đi so sánh \(\frac{2017.2018+1}{2017.2018}\)với\(\frac{2018.2019+1}{2018.2019}\)có :

\(\frac{2017.2018+1}{2017.2018}=\frac{2017.2018}{2017.2018}+\frac{1}{2017.2018}=1+\frac{1}{2017.2018}\left(\cdot\right)\)

\(\frac{2018.2019+1}{2018.2019}=\frac{2018.2019}{2018.2019}+\frac{1}{2018.2019}\left(\cdot\cdot\right)\)

\(\frac{1}{2017.2018}>\frac{1}{2018.2019}\left(\cdot\cdot\cdot\right)\)Từ \(\left(\cdot\right);\left(\cdot\cdot\right)\&\left(\cdot\cdot\cdot\right)\Rightarrow\frac{2017.2018+1}{2017.2018}>\frac{2018.2019+1}{2018.2019}\)

\(\Leftrightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}.\)

#)Trả lời :

\(\frac{2017\times2018}{2017\times2018+1}=\frac{0}{1}=0\)

\(\frac{2018\times2019}{2018\times2019+1}=\frac{0}{1}=0\)

\(\Rightarrow\frac{2017\times2018}{2017\times2018+1}=\frac{2018\times2019}{2018\times2019+1}\)

Ta có:

\(\frac{2017.2019}{2018.2018}\)

\(=\frac{2017.\left(2018+1\right)}{\left(2017+1\right).2018}\)

\(=\frac{2017.2018+2017}{2017.2018+2018}\)

Vì \(2017.2018+2017< 2017.2018+2018\)( tử nhỏ hơn mẫu )

\(\Rightarrow\frac{2017.2018+2017}{2017.2018+2018}< 1\)

Vậy \(\frac{2017.2019}{2018.2018}< 1\)

        ( Mk nghĩ vậy )

                          ~~~~~~~Hok tốt~~~~~~~

\(\frac{2017.2019}{2018.2018}=\frac{2017.\left(2018+1\right)}{2018.\left(2017+1\right)}=\frac{2017.2018+2017}{2018.2017+2018}\)

\(2017< 2018\Rightarrow2017.2018+2017< 2018.2017+2018\Rightarrow\frac{2017.2018+2017}{2018.2017+2018}< 1\Rightarrow\frac{2017.2019}{2018.2018}< 1\)

a) ta có: \(A=\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)

\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)

\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)

\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

=> A < B

a)A= 2017*2018/2017*2018-1/2017*2018=1-1/2017*2018

    B = 2018*2019/2018*2019-1/2018*2019=1-1/2018*2019

vì 1/2017*2018>1/2018*2019=> A<B

b)

ta có: \(Q=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)

\(Q=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+1\)

\(Q=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)

\(Q=50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{P}{Q}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{50.\left(\frac{1}{2}+...+\frac{1}{47}+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}\right)}=\frac{1}{50}\)

cái gì đấy

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)

b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2\left(1-\frac{1}{2019}\right)\)

\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)

\(=2.\frac{2018}{2019}\)

\(=\frac{4036}{2019}\)

Phần c tương tự nha

a) \(\frac{1}{1.2}\) +  \(\frac{1}{2.3}\) + .......+  \(\frac{1}{2017.2018}\)

= 1 -  \(\frac{1}{2}\) + \(\frac{1}{2}\) -  \(\frac{1}{3}\) + .......+  \(\frac{1}{2017}\) -   \(\frac{1}{2018}\)

= 1 -  \(\frac{1}{2018}\) =  \(\frac{2017}{2018}\)

câu a) mik sửa đề một tí ko biết có đúng ko

câu b , c tương tự nhưng cần lấy tử ra chung 

a)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b)nhóm 2 ra ngoài rồi làm như câu a

c)nhóm 4 ra rồi làm như câu a

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Đặt \(2016=a\) biểu thức trên trở thành:

\(P=\dfrac{\left(a^2\left(a+10\right)+31\left(a+1\right)-1\right)\left(a\left(a+5\right)+4\right)}{\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)}=\dfrac{A}{B}\)

Với \(B=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)\)

Ta có: \(a^2\left(a+10\right)+31\left(a+1\right)-1=a^3+10a^2+31a+30\)

\(=a^3+5a^2+6a+5a^2+25a+30=a\left(a^2+5a+6\right)+5\left(a^2+5a+6\right)\)

\(=\left(a+5\right)\left(a^2+5a+6\right)=\left(a+5\right)\left(a^2+2a+3a+6\right)\)

\(=\left(a+5\right)\left(a+2\right)\left(a+3\right)\)

Và \(a\left(a+5\right)+4=a^2+5a+4=a^2+a+4a+4=\left(a+1\right)\left(a+4\right)\)

\(\Rightarrow A=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)=B\)

\(\Rightarrow P=\dfrac{A}{B}=1\)

Đề là gì vậy bạn !

\(\frac{2017.2019}{2018.2019}\)< 1

Tk nha!!

\(Vì\) \(\frac{2017}{2018}< 1\)mà \(\frac{2019}{2018}>1\)nên

\(\Rightarrow\frac{2017}{2018}< \frac{2019}{2018}\)

2017.2019/2018.2018 <1 nha

k mk nha !