cho so A=\(\frac{2013+\frac{1}{2}}{\left(2012+\frac{1}{2}\right)^2+2013+\frac{1}{2}}\)
B=\(\frac{2013+\frac{1}{3}}{\left(2012+\frac{1}{3}\right)^2+2013+\frac{1}{3}}\)
so sanh A va B
a , | 3 - 2x | = x + 1
b , \(\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2014}\right).x=\frac{2013}{1}+\frac{2012}{2}+......+\frac{2}{2012}+\frac{1}{2013}\)
a, ĐK: \(x+1\ge0\Leftrightarrow x\ge-1\)
Ta có: |3-2x|=x+1
=>\(\orbr{\begin{cases}3-2x=x+1\\3-2x=-x-1\end{cases}\Rightarrow\orbr{\begin{cases}x+2x=3-1\\-x+2x=3+1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=2\\x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\left(tmđk\right)\\x=4\left(tmđk\right)\end{cases}}}\)
Vậy x=2/3 hoặc x=4
b, Xét VP ta có: \(\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}+1\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(x=\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}=2014\)
Vậy x=2014
So sánh A và B,biết:
\(A=\left(1+\frac{1}{2013}\right)\left(1+\frac{1}{2013^2}\right).....\left(1+\frac{1}{2013^n}\right)\) (với n là số nguyên dương)
\(B=\frac{2013^2-1}{2012^2-1}\)
học trước chương trình ak, mk chưa học đn dạng này
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
Tìm x biết : \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\left(\frac{2012}{2}+1\right)+...+\left(\frac{2}{2012}+1\right)+\left(\frac{1}{2013}+1\right)+1\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2014}{2}+...+\frac{2014}{2012}+\frac{2014}{2013}+\frac{2014}{2014}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)\)
\(x=\frac{2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)
\(x=2014\)
Cho \(A=\left(\frac{1}{^{2^2}}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)......\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)va\)\(B=-\frac{1}{2}.\)Hay so sanh A va B
(\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
tìm x
Ta có: \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=1+\left(1+\frac{2012}{2}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=\frac{2014}{2014}+\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
\(\Rightarrow x=2014\)
Lưu ý: số 2013 ở dòng T2 được tách ra làm 2013 số 1
Cho A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) va B = \(\frac{-1}{2}\), So sanh A va B
A = \(-\frac{1.3}{2.2}.-\frac{2.4}{3.3}.\cdot\cdot\cdot-\frac{2013.2015}{2014.2014}=-\frac{\left(1.2.3...2013\right).\left(3.4.5....2015\right)}{\left(2.3....2014\right).\left(2.3....2014\right)}=-\frac{2.2015}{2014}=-\frac{4030}{2014}
Câu 1: Rút gọn: \(A=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)
Câu 2: Cho \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\). Tính \(\left(S-P\right)^{2013}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right).x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)Tìm x biết: