Question

[DPKhanh]

\(\dfrac{x+1}{124}\)+\(\dfrac{x+2}{123}\)=\(\dfrac{x+3}{122}\)+\(\dfrac{x+4}{121}\)

Answer 1

[Nguyễn Huy Tú]

\(\dfrac{x+1}{124}+1+\dfrac{x+2}{123}+1=\dfrac{x+3}{122}+1+\dfrac{x+4}{121}+1\)

\(\Leftrightarrow\dfrac{x+125}{124}+\dfrac{x+125}{123}=\dfrac{x+125}{122}+\dfrac{x+125}{121}\)

\(\Leftrightarrow\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\ne0\right)=0\Leftrightarrow x=-125\)

Answer 2

[Trần Đức Huy]

<=>\(\dfrac{x+1}{124}+\dfrac{x+2}{123}-\dfrac{x+3}{122}-\dfrac{x+4}{121}=0\)

<=>\(\left(\dfrac{x+1}{124}+1\right)+\left(\dfrac{x+2}{123}+1\right)-\left(\dfrac{x+3}{122}+1\right)-\left(\dfrac{x+4}{121}+1\right)=0\)

<=>\(\dfrac{x+125}{124}+\dfrac{x+125}{123}-\dfrac{x+125}{122}-\dfrac{x+125}{121}=0\)

<=>\(\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\right)=0\)

<=>x+125=0

<=>x=-125