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Giải:
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(\Leftrightarrow2+\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=2+\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(\Leftrightarrow1+\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}=1+\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}\)
\(\Leftrightarrow\left(1+\dfrac{x+1}{2015}\right)+\left(1+\dfrac{x+2}{2014}\right)=\left(1+\dfrac{x+3}{2013}\right)+\left(1+\dfrac{x+4}{2012}\right)\)
\(\Leftrightarrow\dfrac{x+1+2015}{2015}+\dfrac{x+2+2014}{2014}=\dfrac{x+3+2013}{2013}+\dfrac{x+4+2012}{2012}\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
Vì \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)
Nên \(x+2016=0\)
\(\Leftrightarrow x=0-2016\)
\(\Leftrightarrow x=-2016\)
Vậy ...
Chúc bạn học tốt!
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(\Rightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}+1\)
\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)
\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)
\(\Rightarrow\left(x+2016\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
do \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)
\(\Rightarrow x+2016=0\Rightarrow x=2016\)
váy x=2016
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}+\dfrac{x+4}{2012}+\dfrac{x+2024}{2}=0\)
\(\Leftrightarrow(\dfrac{x+1}{2015}+1)+(\dfrac{x+2}{2014}+1)+(\dfrac{x+3}{2013}+1)+(\dfrac{x+4}{2012}+1)+\dfrac{x+2024}{2}-4=0\)\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}+\dfrac{x+2016}{2}=0\)\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}\right)=0\)
Hiển nhiên: \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}>0\)
\(\Leftrightarrow x+2016=0\Leftrightarrow x=-2016\)
b) \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{1}{18}\\< =>\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ < =>\dfrac{1}{x+1}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ quyđồngmẫuvàkhửmẫu\\ x^{2^{ }}+6x-27=0\\ giảipttìmđược:x=3;x=-9\)
a) \(\frac{x-2015}{1}+\frac{x-2014}{2}+\frac{x-2013}{3}+...+\frac{x-1}{2015}+\frac{x}{2016}=0\\ \Leftrightarrow\frac{x-2015}{1}-1+\frac{x-2014}{2}-1+...+\frac{x-1}{2015}-1+\frac{x}{2016}-1=-2016\)
\(\Leftrightarrow\frac{\left(x-2016\right).1}{1}+\frac{\left(x-2016\right).1}{2}+\frac{\left(x-2016\right).1}{3}+...+\frac{\left(x-2016\right).1}{2015}+\frac{\left(x-2016\right).1}{2016}=-2016\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=-2016\)
tới đây mình chịu. mình nghĩ là phương trình bạn cho là bằng 2016 chứ, như thế giải mới được, còn như này thì mình bó tay
b)
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\\ \Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x+12-32=0\\ \Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-10\\x=2\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={-10;2}
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(pt\Leftrightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}+1\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
Dễ thấy: \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)
\(\Rightarrow x+2016=0\Rightarrow x=-2016\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(\dfrac{x+1}{2015}+1+\dfrac{x+1}{2014}+1-\dfrac{x+3}{2013}-1-\dfrac{x+4}{2012}-1=0\)
\(\dfrac{x+1+2015}{2015}+\dfrac{x+2+2014}{2014}-\dfrac{x+3+2013}{2013}-\dfrac{x+4+2012}{2012}=0\)
\(\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)
\(\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
Vì \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}< 0\)
Nên để:\(\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
Thì \(x+2016=0\Leftrightarrow x=-2016\)
\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)
\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)
\(\frac{x-2010}{2010}+\frac{x-2010}{2011}+\frac{x-2010}{2012}+\frac{x-2010}{2013}+\frac{x-2010}{2014}=0\)
\(\left(x-2010\right).\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)=0\)
mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)
Vậy x=-2010
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}+\dfrac{x+3}{2010}=\dfrac{x-1}{2014}+\dfrac{x-2}{2015}+\dfrac{x-3}{2016}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2012}+1\right)+\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+3}{2010}\right)=\left(\dfrac{x-1}{2014}+1\right)+\left(\dfrac{x-2}{2015}+1\right)+\left(\dfrac{x-3}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2014}-\dfrac{x+2013}{2015}-\dfrac{x+2013}{2016}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
Tìm x:
\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)
\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2010}-1\right)+\left(\dfrac{x+1}{2011}-1\right)+\left(\dfrac{x+2}{2012}-1\right)+\left(\dfrac{x+3}{2013}-1\right)+\left(\dfrac{x+4}{2014}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{2010}+\dfrac{x-2010}{2011}+\dfrac{x-2010}{2012}+\dfrac{x-2010}{2013}+\dfrac{x-2010}{2014}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)=0\)
\(\Leftrightarrow x=2010\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(\Leftrightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}+1\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
Mà \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)
\(\Leftrightarrow x+2016=0\Leftrightarrow x=-2016\)
Vậy x = -2016
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
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
<=>\(\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
<=>\(\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
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
\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)
\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)
\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)
\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)