Question

BT1: Tìm x, biết:

6) \(\dfrac{x+1}{2017}+\dfrac{x+2}{2016}=\dfrac{x+3}{2015}-1\)

Answer 1

\(\dfrac{x+1}{2017}+\dfrac{x+2}{2016}=\dfrac{x+3}{2015}-1\)

\(\Leftrightarrow\dfrac{x+1}{2017}+1+\dfrac{x+2}{2016}+1=\dfrac{x+3}{2015}-1+2\)

\(\Leftrightarrow\dfrac{x+100}{2017}+\dfrac{x+100}{2016}=\dfrac{x+100}{2015}\)

\(\Leftrightarrow\dfrac{x+100}{2017}+\dfrac{x+100}{2016}+\dfrac{x+100}{2015}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)

Do \(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\ne0\) nên \(x+100=0\)

\(\Leftrightarrow x=\left(-100\right)\)

Vậy \(x=\left(-100\right)\)

Answer 2

(Thêm ở cả 2 vế cùng một số để tạo ra nhân tử chung ở tử (x + 2018))