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\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(1-\frac{2016}{1}\right)+\left(1-\frac{2017}{2}\right)+...+\left(1-\frac{4030}{2015}\right)\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
\(\Rightarrow x=2015\)
Không hiểu thì hỏi mình nhé! Thiên dâng bữa nay chăm chỉ đột xuất ta???
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015.\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\)
=> x = 2015
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x+2015=\frac{2016}{1}+\frac{2017}{2}+\frac{2018}{3}+...+\frac{4030}{2015}\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}=2015.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)\(\Rightarrow x=2015\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4030}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
\(\Rightarrow x=2015\)
Bạn có thể tham khảo nhé!^-^