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\(\frac{x-4}{2015}-\frac{1}{2015}=\frac{10-2x}{2015}\)
\(\Rightarrow\frac{x-4}{2015}-\frac{10-2x}{2015}=\frac{1}{2015}\)
\(\Rightarrow\frac{x-4-\left(10-2x\right)}{2015}=\frac{1}{2015}\)
\(\Rightarrow\frac{\left(x+2x\right)-\left(4+10\right)}{2015}=\frac{1}{2015}\)
\(\Rightarrow\frac{3x-14}{2015}=\frac{1}{2015}\)
\(\Rightarrow\left(3x-14\right).2015=2015\)
\(\Rightarrow3x-14=1\) ( bớt cả 2 vế đi 2015 lần )
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
https://dethi.violet.vn/present/showprint/entry_id/11072330
bạn vào link trên sẽ có full đề và đáp án
p/s: nhớ k cho mình nha <3
\(\frac{x-2}{4}=-\frac{16}{2-x}\)
\(\Leftrightarrow\frac{x-2}{4}=\frac{16}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=4.16=64\)
\(\Leftrightarrow\left(x-2\right)^2=8^2\)
\(\Leftrightarrow\left(x-2-8\right)\left(x-2+8\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+6\right)=0\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)
\(\frac{x-1}{2015}-\frac{1}{2015}=\frac{10-2x}{2015}\)
\(\Rightarrow x-1-1=10-2x\)
\(\Rightarrow x-2=10-2x\)
\(\Rightarrow-2x+x=10+2\)
\(\Rightarrow-x=12\Rightarrow x=-12\)
Tìm x, biết:
\(\frac{x+1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}=\frac{x-1}{2014}+\frac{x-1}{2015}\)
\(\frac{x-1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}=\frac{x-1}{2014}+\frac{x-1}{2015}\)
\(\Rightarrow\frac{x-1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}-\frac{x-1}{2014}-\frac{x-1}{2015}=0\)
\(\left(x-1\right).\left(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
mà \(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
=> x - 1 = 0
x = 1
bn có chép sai đề ko z???
\(\frac{x+2}{2013}+\frac{x+1}{2014}=\frac{x}{2015}+\frac{x-1}{2016}\)
\(\Leftrightarrow\)\(\frac{x+2}{2013}+1+\frac{x+1}{2014}+1=\frac{x}{2015}+1+\frac{x-1}{2016}+1\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2014}=\frac{x+2015}{2015}+\frac{x+2015}{2016}\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
Do\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}>0\)
=>x+2015=0
<=>x=-2015
=> \(\frac{x+2015-2013}{2013}+\frac{x+2015-2014}{2014}=\frac{x+2015-2015}{2015}+\frac{x+2015-2016}{2016}\)
<=> \(\frac{x+2015}{2013}-1+\frac{x+2015}{2014}-1=\frac{x+2015}{2015}-1+\frac{x+2015}{2016}-1\)
<=> \(\frac{x+2015}{2013}+\frac{x+2015}{2014}-\frac{x+2015}{2015}-\frac{x+2015}{2016}=0\)
<=> \(\left(x+2015\right).\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
<=> x + 2015 = 0 Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
<=> x = -2015
\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)
\(\Rightarrow\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}-\left(\frac{x+2016}{2014}+\frac{x+2016}{2015}\right)=0\)
\(\Rightarrow\left(x+2016\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
Ta có : \(\frac{x+1}{2015}=\frac{2015}{x+1}\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)=2015.2015\)
\(\Rightarrow\left(x+1\right)^2=2015^2\)
\(\Rightarrow x+1=\pm2015\)
\(\Rightarrow\orbr{\begin{cases}x+1=2015\\x+1=-2015\end{cases}\Rightarrow\orbr{\begin{cases}x=2014\\x=-2016\end{cases}}}\)
Vậy x = 2014 và x = -2016