1.Tìm x và y biết rằng 2x+242=3y
2.Thực hiện phép tính 1.3+2.4+3.5+...+99.100
thực hiện phép tính
a) A= 1.3+2.4+3.5+....+97.98+99.100
b) B=1.100+2.99+3.98+.....+100.1
c) C= 1.1+2.2+....+100.100
thực hiện phép tính : (1+ 1/1.3) ( 1+1/2.4) ( 1+1/3.5)...( 1+1/2020.2021)
Thực hiện phép tính 2. (1/1.3 + 1/3.5+ 1/ 5.7 + .... + 1/ 99.100) ta được kết quả:
Thực hiện phép tính:
\(\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{2019.2021}\right)\)
Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)
\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)
Thực hiện phép tính:
\(\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{2019.2021}\right)\)
Thực hiện phép tính: \(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)
\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
2*(1/1*3+1/3*5+.......+1/99*100)
=2*(2/1*3+2/3*5+.....+2/99*100)*1/2
=1/3-1/5+1/5-1/7+....+1/99-1/100
=1/3-1/100
=100/300-3/300
=97/300
2(1/1.3+1/3.5+1/5.7+...+1/99.100)
=2/1.3+2/3.5+2/5.7+...+2/99.100
=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/100
=1-1/100
=99/100
Thực hiện phép tính:
A=1.99+2.98+3.97+...+98.2+99.1
B=1.2.3+2.3.4+3.4.5+...+17.28.19
C=1.4+2.5+3.6+...+100.103
D=1.3+2.4+3.5+...+97.99+98.100
Thực hiện phép tính:
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{20.22}\right)\)
Thực hiện phép tính :
A=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)