Question

[đỗ thị kiều trinh]

tìm x biết:

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right).503x\)\(=\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)

Answer 1

[đỗ thị kiều trinh]

thêm số hạng 1 vào bên phải nha

Answer 2

[Nguyễn Thành Trương]

Xét \( A = 1 + \dfrac{{2014}}{2} + \dfrac{{2015}}{3} + ... + \dfrac{{4023}}{{2011}} + \dfrac{{4024}}{{2012}}\\ \)

\(\Rightarrow A - 2012 = \left( {\dfrac{{2014}}{2} - 1} \right) + \left( {\dfrac{{2015}}{3} - 1} \right) + ... + \left( {\dfrac{{4024}}{{2012}} - 1} \right)\\ \Rightarrow A - 2012 = \dfrac{{2012}}{2} + \dfrac{{2012}}{3} + ... + \dfrac{{2012}}{{2012}}\\ \Rightarrow A - 2012 = 2012\left( {\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow A = 2012\left( {1 + \dfrac{1}{2} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)503x = 2012\left( {1 + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow x = \dfrac{{2012}}{{503}} = 4 \)