cái gì đấy

\(\frac{2019.2020-4038}{2017.2019+2019}\)

\(=\frac{2019.2020-2.2019}{2019\left(2017+1\right)}=\frac{2019\left(2020-2\right)}{2019.2018}=\frac{2019.2018}{2019.2018}=1\)

\(A=\frac{2019.2020-4038}{2017.2019+2019}\)

   \(=\frac{2019\left(2020-2\right)}{2019\left(2017+1\right)}\)

   \(=\frac{2019.2018}{2019.2018}=1\)

Vậy \(A=1.\)

Mà lớp 5 làm gì đã học đến dấu \(.\)(dấu nhân lớp 5 viết kiểu này cơ: x )

Chúc em học tốt.

\(A=\frac{2019.2020-4038}{2017.2019+2019}\)

\(A=\frac{2019.2020-2.2019}{2017.2019+2019}\)

\(A=\frac{2019.\left(2020-2\right)}{2019.\left(2017+1\right)}\)

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

\(A=1\)

Tham khảo nhé~

d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)

\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)

\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)

\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)

\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)

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

\(=-1\)

c)\(\frac{2929-101}{2.1919+404}=\frac{29.101-101}{29.101+101.4}=\frac{101\left(29-1\right)}{101\left(29+4\right)}=\frac{28}{33}\)

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

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

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

\(\frac{2015+2016.2017}{2017.2018-2019}\)

\(=\frac{2015+2016}{2018-2019}\)

\(=\frac{4031}{-1}\)

\(=-4031\)

(Mik làm bừa thôi bạn, sai đừng k sai nha :( Tội mik lắm)

A = 

A = \(1-\frac{1}{2018}\)

A = \(\frac{2017}{2018}\)

Có : 

2.B = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)

2.B = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

2.B = \(1-\frac{1}{2017}\)

2.B = \(\frac{2016}{2017}\)

B = \(\frac{2016}{2017}:2=\frac{1008}{2017}\)

Có :

3.C = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2017.2020}\)

3.C = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\)

3.C = \(\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

C = \(\frac{2019}{2020}:3=\frac{673}{2020}\)

a=1/1-1/2+1/2-1/3+...+1/2017-1/2018

=1/1-1/2018

=kq

may bai duoi lam tuong tu nha

mình chưa điền kết quả ban tu dien nha 

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

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

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

\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2015\cdot2017}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2015\cdot2017}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2017}\right)\)

\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)

\(B=\frac{1008}{2017}\)

\(C=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{2017\cdot2020}\)

\(C=\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{2017\cdot2020}\right)\)

\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\right)\)

\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2020}\right)\)

\(C=\frac{1}{3}\cdot\frac{2019}{2020}\)

\(C=\frac{673}{2020}\)