tìm x thỏa mãn
x/2+x/4+x/2014=x/3+x/5+x/2015
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\(\frac{x+3}{2013}+1+\)\(\frac{x+4}{2012}+1+\frac{x+5}{2011}+1\)=\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1+\frac{x}{2016}+1\)
\(\Rightarrow\frac{x+2016}{2013}+\frac{x+2016}{2012}+\frac{x+2016}{2011}=\frac{x+2016}{2014}+\frac{x+2016}{2016}\)
\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2016}=0\right)\)
Vì 1/2016+...+1/2011>0 nên (x+2016)=0
suy ra x= -2016
nếu đúng xin kết bạn
ch
=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2
=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)
Với x+2020=0=>x=-2020
Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí
Vậy x=-2020
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)
\(\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\left(x+2016\right).\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(x+2016=0\)
\(x=-2016\)
a: x<50
mà x tròn chục
nên \(x\in\left\{10;20;30;40\right\}\)
b: 33<x<77
mà x tròn chục
nên \(x\in\left\{40;50;60;70\right\}\)
Vì \(\frac{x}{2}+\frac{x}{4}+\frac{x}{2014}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2015}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2014}\right)=x\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2015}\right)\)
Mà \(\frac{1}{2}+\frac{1}{4}+\frac{1}{2014}\ne\frac{1}{3}+\frac{1}{5}+\frac{1}{2015}\)
\(\Rightarrow x=0\)
\(X=0\)