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Question

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$\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1$

trần thảo lê · Accepted Answer

$\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1$

$\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-1}{1008}-2\right)+\left(\dfrac{x}{2017}-1\right)$

$\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{1008}-\dfrac{x-2017}{2017}=0$

$\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0$

$x-2017=0$       vì$\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0$

$\Rightarrow x=2017$Câu trả lời: $\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1$

$\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-1}{1008}-2\right)+\left(\dfrac{x}{2017}-1\right)$

$\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{1008}-\dfrac{x-2017}{2017}=0$

$\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0$

$x-2017=0$       vì$\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0$

$\Rightarrow x=2017$

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