Rút gọn Hộ mk với!
\(\frac{2019\cdot2020-4040}{2017\cdot2018+4034}\)
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\(\frac{2019.2020-4038}{2017.2019+2019}\)
\(=\frac{2019.2020-2.2019}{2019\left(2017+1\right)}=\frac{2019\left(2020-2\right)}{2019.2018}=\frac{2019.2018}{2019.2018}=1\)
\(A=\frac{2019.2020-4038}{2017.2019+2019}\)
\(=\frac{2019\left(2020-2\right)}{2019\left(2017+1\right)}\)
\(=\frac{2019.2018}{2019.2018}=1\)
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.
d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)
\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)
\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)
\(=-\frac{2018}{2018}\)
\(=-1\)
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 = \(\frac{2019}{2020}:3=\frac{673}{2020}\)
\(\frac{2019.2020-4040}{2017.2018+4034}\)=\(\frac{\left(2017+2\right).2020-4040}{2017.2018+2017.2}\)
=\(\frac{2017.2020+2.2020-4040}{2017.\left(2018+2\right)}\)
=\(\frac{2017.2020+4040-4040}{2017.2020}\)
=\(\frac{2017.2020+0}{2017.2020}\)
=\(\frac{1}{1}\)=1