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\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-1}{1008}-2\right)+\left(\dfrac{x}{2017}-1\right)\)
\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{1008}-\dfrac{x-2017}{2017}=0\)
\(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)
\(x-2017=0\) vì\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0\)
\(\Rightarrow x=2017\)
\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)
\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)
\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)
\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}+\dfrac{x+4}{2012}+\dfrac{x+2024}{2}=0\)
\(\Leftrightarrow(\dfrac{x+1}{2015}+1)+(\dfrac{x+2}{2014}+1)+(\dfrac{x+3}{2013}+1)+(\dfrac{x+4}{2012}+1)+\dfrac{x+2024}{2}-4=0\)\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}+\dfrac{x+2016}{2}=0\)\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}\right)=0\)
Hiển nhiên: \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}>0\)
\(\Leftrightarrow x+2016=0\Leftrightarrow x=-2016\)
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
a: \(\Leftrightarrow x+2016=0\)
hay x=-2016
b: \(\Leftrightarrow x-100=0\)
hay x=100
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\Leftrightarrow\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{1008}-2+\dfrac{x}{2017}-1\) \(\Leftrightarrow\dfrac{x-3-2014}{2014}+\dfrac{x-2-2015}{2015}=\dfrac{x-1-2016}{1008}-\dfrac{x-2017}{2017}\) \(\Leftrightarrow\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{1008}+\dfrac{x-2017}{2017}\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)
Vì: \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0\)
Suy ra: x -2017 = 0
=> x = 2017
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
⇔ \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{2008}-2+\dfrac{x}{2017}-1\)
⇔\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2008}+\dfrac{x-2017}{2017}\)
⇔\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2008}-\dfrac{x-2017}{2017}=0\)
⇔\(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2008}-\dfrac{1}{2017}\right)=0\)
⇔x-2017=0
⇔x=2017
vậy phương trình có tập nghiệm là S={2017}
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 \(x+2016=0\Leftrightarrow x=-2016\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(\Leftrightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}+1\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
Mà \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)
\(\Leftrightarrow x+2016=0\Leftrightarrow x=-2016\)
Vậy x = -2016
\(\Leftrightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x}{1008}+\dfrac{x+3}{2013}+1\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x}{1008}+\dfrac{x+2016}{2013}\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x}{1008}-\dfrac{x+2016}{2013}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(-\dfrac{x}{1008}+\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x}{1008}+1+\frac{x+3}{2013}\)
\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+1008}{1008}+1+\frac{x+3}{2013}+1\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{1008}+\frac{x+2016}{2013}\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}-\frac{1}{2013}\right)=0\)
vì \(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}+\frac{1}{2013}\right)\ne0\)nên
x+2016=0\(\Leftrightarrow\)x=-2016