1,tìm x:\(\dfrac{2-x}{2013}-1=\dfrac{1-x}{2014}-\dfrac{x}{2015}\)
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\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)
\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)
\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)
\(\Leftrightarrow x=2017\) \(\text{Vậy }x=2017\)
\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)
\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)
\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)
\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)
\(\Leftrightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x}{1008}+\dfrac{x+3}{2013}+1\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x}{1008}+\dfrac{x+2016}{2013}\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x}{1008}-\dfrac{x+2016}{2013}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(-\dfrac{x}{1008}+\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x}{1008}+1+\frac{x+3}{2013}\)
\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+1008}{1008}+1+\frac{x+3}{2013}+1\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{1008}+\frac{x+2016}{2013}\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}-\frac{1}{2013}\right)=0\)
vì \(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}+\frac{1}{2013}\right)\ne0\)nên
x+2016=0\(\Leftrightarrow\)x=-2016
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}=-3\)
\(\left(\dfrac{x+1}{2015}+1\right)+\left(\dfrac{x+2}{2014}+1\right)+\left(\dfrac{x+3}{2013}+1\right)=0\)
\(\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}=0\)
\(\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)
\(\Rightarrow x+2016=0\Rightarrow x=-2016\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}=-3\)
\(\Rightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1+\dfrac{x+3}{2013}+1=0\)
\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}=0\)
\(\Rightarrow\left(x+2016\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)
\(\Rightarrow x+2016=0\Rightarrow x=-2016\)
Chúc bạn học tốt!!!
\(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)
\(\Leftrightarrow\dfrac{x+4}{2012}+1+\dfrac{x+3}{2013}+1=\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}\)
\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}=\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\)
\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}-\left(\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\right)=0\)
\(\Leftrightarrow x+2016.\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)
Vì \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
Vậy \(x=-2016\) tại biểu thức \(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)
Theo đề ta có: x+4/2012+x+3/2013=x+2/2014+x+1/2015
=>x+4/2012+x+3/2013-x+2/2014+x+1/2015=0
=>( x+4/2012+1)+(x+3/2013+1)-(x+2/2014+1)+(x+1/2015+1)
=>x+2016/2012+x+2016/2013-x+2016/2014-x+2016/2015=0
=>x+2016.(1/2012+1/2013-1/2014-1/2015)=0
Do 1/2012+1/2013-1/2014-1/2015>0
nên x+2016=0
=>x=-2016
Vậy 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: \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}>0\)
\(\Leftrightarrow x+2016=0\Leftrightarrow x=-2016\)
\(\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)
<=>\(\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)
<=>(x+2017)(1/2016+1/2015-1/2014-1/2013)=0
vì 1/2016+1/2015-1/2014-1/2013 khác 0
nên x+2017=0<=>x=-2017
vậy................
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