Question

[Đặng Anh Thư_Thư Đặng-A1]

Bài 1:so sánh: 2017/2018+2018/2019 và ( 2017+2018/2018/2019)

 Bài 2: (1/2003+1/2004+1/2005)/(2/2003+2/2004+2/2005)

 Bài 3: 2013+ (2013/1+2)+(2013/1+2+30+...+(2013/1+2+3+..+2012)

Answer 1

[ミ★๖ۣۜSεηαbĭ ๖ۣۜTĭʂт ๖ۣۜKεү ★彡]

Bài 1

\(\frac{2017}{2018}+\frac{2018}{2019}\)và \(\left(\frac{2017+2018}{2018+2019}\right)\)mk chữa lại đề luôn đó 

Ta tách :

\(\frac{2017}{\left(2018+2019\right)+2018}\)

đến đây ta tách 

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

vậy....

mấy câu khác tương tự 

Answer 2

[Xyz OLM]

2) \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{2}{2003}+\frac{2}{2004}+\frac{2}{2005}}\)

= \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{2.\frac{1}{2003}+2.\frac{1}{2004}+2.\frac{1}{2005}}\)

=\(\frac{1\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}{2.\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}\)

= \(\frac{1}{2}\)

3) \(2013+\left(\frac{2013}{1+2}\right)+\left(\frac{2013}{1+2+3}\right)+...+\left(\frac{2013}{1+2+3+...+2012}\right)\)

= \(2013.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\right)\)

= \(2013.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2025078}\right)\)

= \(2013.2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4050156}\right)\)

=\(4026.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)

= \(4026.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)

= \(4026.\left(1-\frac{1}{2013}\right)\)

= \(4026.\frac{2012}{2013}\)

=\(4024\)