\(\frac{1-x}{2015}+\frac{2-x}{1007}+\frac{3-x}{671}=\frac{1992-x}{4}\)
\(\left(\frac{1-x}{2015}+1\right)+\left(\frac{2-x}{1007}+2\right)+\left(\frac{3-x}{671}+3\right)=\frac{1992-x}{4}+6\)
\(\left(\frac{2016-x}{2015}+\frac{2016-x}{1007}+\frac{2016-x}{671}\right)=\frac{2016-x}{4}\)
\(\Leftrightarrow\left(2016×x\right)×\left(\frac{1}{2015}+\frac{1}{1007}+\frac{1}{671}-\frac{1}{4}\right)=0\)
Vì\(\frac{1}{2015}+\frac{1}{1007}+\frac{1}{671}-\frac{1}{4}\ne0\)
\(\Rightarrow2016-x=0\)
\(\Rightarrow x=2016\)
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