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\(A=\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+...+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{1+\left(\dfrac{1}{2013}+1\right)+\left(\dfrac{2}{2012}+1\right)+\left(\dfrac{3}{2011}+1\right)+...+\left(\dfrac{2012}{2}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{\dfrac{2014}{2014}+\dfrac{204}{2013}+\dfrac{2014}{2012}+\dfrac{2014}{2011}+...+\dfrac{2014}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}=2014\)
mình ko chắc đúng nha !
Số số hạng của tử là :
(2013-1):1+1=2013(số hạng)
\(\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+.....+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=\dfrac{\dfrac{1}{2013}+1+\dfrac{2}{2012}+1+....+\dfrac{2012}{2}+1+\dfrac{2013}{1}-2012}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=\dfrac{\dfrac{2014}{2013}+\dfrac{2014}{2012}+....+\dfrac{2014}{2}+1}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=2014\left(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\right)\)
=2014
Mình ghi thêm ở cái dâu bằng thứ 2 cuối cùng trên tử có ghi trừ 2012 là do tử có 2013 hạng tử mà mình chỉ cộng 1 cho 2012 hạng tử nên phải trừ đi 2012
Đặt \(S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+....+\frac{2013}{1+2013^2+2013^4}\)
Xét:
\(\frac{k}{k+k^2+k^4}=\frac{1}{2}\cdot\frac{k^2+k+1-k^2+k-1}{k^4+k^2+1}\)
\(=\frac{1}{2}\cdot\frac{k\left(k+1\right)+1-k\left(k-1\right)-1}{\left(k^2+1\right)^2-k^2}\)
\(=\frac{1}{2}\left[\frac{1}{k\left(k-1\right)+1}-\frac{1}{k\left(k+1\right)+1}\right]\)
Áp dụng :
\(S=\frac{1}{2}\left[\frac{1}{1\cdot0+1}-\frac{1}{1\cdot2+1}+\frac{1}{2\cdot1+1}-\frac{1}{2\cdot3+1}+.....+\frac{1}{2013\cdot2012+1}-\frac{1}{2013\cdot2014+1}\right]\)
\(=\frac{2027091}{4054183}\)
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
<=>\(\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
<=>\(\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
vì 1/2013+1/2012+1/2011-1/2010-1/2009-1/2008 khác 0
=>x-2014=0<=>x=2014
bạn hiểu chứ?
Xuyên Cúc: -1 tại vì còn phải tùy bài, mk phải làm thế nào để tử giống nhau, thì có trường hợp + có trường hợp -, cái đấy còn tùy
còn 1/2013...+... khác 0 vì chắc chắn nó sẽ khác 0, cái dãy số đấy k có chuyện bằng 0 đc , tớ cũng chả biết giải thích thế nào nữa == bt nếu làm ra như vầy : (x+1)(1/2+...+..) thì x+1=0 còn cái vế còn lại sẽ khác 0, hầu như là thế chứ tớ chưa thấy trường hợp nào mà vế x+1 khác 0 còn vế kia bằng 0 cả
Ta có : \(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+\dfrac{x+3}{2012}+\dfrac{x+4}{2011}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+\left(\dfrac{x+3}{2012}+1\right)+\left(\dfrac{x+4}{2011}+1\right)=4\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}=4\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}\right)=4\) \(\Leftrightarrow\left(x+2015\right).0,002=4\) ( mik lấy gần bằng nha )
\(\Leftrightarrow x+2015=2000\Leftrightarrow x=-15\)
Vậy phương trình có nghiệm là x=-15
a/ Đặt \(x^2+x+1=a\Rightarrow x^2+x+2=a+1\)
Pt trở thành \(a\left(a+1\right)-12=0\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2-3a+4a-12=0\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+4\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+1=3\\x^2+x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
2/ \(\dfrac{x+1}{2014}+1+\dfrac{x+2}{2013}+1=\dfrac{x+3}{2012}+1+\dfrac{x+4}{2011}+1\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}=\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}\)
\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
\(\Leftrightarrow x+2015=0\) (do \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))
\(\Rightarrow x=-2015\)
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
\(\Leftrightarrow\dfrac{5-x^2}{2012}=\dfrac{4-x^2}{2013}+1-\dfrac{x^2-3}{2014}\)
\(\Leftrightarrow\dfrac{5-x^2}{2012}+1=\dfrac{4-x^2}{2013}+1+\dfrac{3-x^2}{2014}+1\)
\(\Leftrightarrow2017-x^2=0\)
hay \(x\in\left\{\sqrt{2017};-\sqrt{2017}\right\}\)
Ta có:
\(1+a^2+a^4=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
Từ đây thì ta có:
\(A=\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+...++\dfrac{2013}{1+2013^2+2013^4}\)
\(\Leftrightarrow2A=\dfrac{2}{\left(1^2-1+1\right)\left(1^2+1+1\right)}+\dfrac{4}{\left(2^2-2+1\right)\left(2^2+2+1\right)}+...+\dfrac{4026}{\left(2013^2-2013+1\right)\left(2013^2+2013+1\right)}\)
\(=\dfrac{2}{1.3}+\dfrac{4}{3.7}+...+\dfrac{4026}{4050157.4054183}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{7}+...+\dfrac{1}{4050157}-\dfrac{1}{4054183}=1-\dfrac{1}{4054183}=\dfrac{4054182}{4054183}\)
\(\Rightarrow A=\dfrac{2027091}{4054183}\)
Thank you bn nhìu nha!!!