Question

\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\)

Answer 1

\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\\ \Leftrightarrow\dfrac{315-x}{101}+1+\dfrac{313-x}{103}+1+\dfrac{311-x}{105}+1+\dfrac{309-x}{107}+1=0\\ \Leftrightarrow\dfrac{416-x}{101}+\dfrac{416-x}{103}+\dfrac{416-x}{105}+\dfrac{416-x}{107}=0\\ \Leftrightarrow\left(416-x\right)\left(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\right)=0\\ \dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}>0\\ \Rightarrow416-x=0\\ \Leftrightarrow x=416\)