Question

\(\text{217-x/207+215-x/209+213-x/211+211-x/213+4=0}\)

Answer 1

\(\frac{217-x}{207}+\frac{215-x}{209}+\frac{213-x}{211}+\frac{211-x}{213}+4=0\)

\(\Leftrightarrow\)\(\left(\frac{217-x}{207}+1\right)+\left(\frac{215-x}{209}+1\right)+\left(\frac{213-x}{211}+1\right)+\left(\frac{211-x}{213}+1\right)=0\)

\(\Leftrightarrow\)\(\frac{424-x}{207}+\frac{424-x}{209}+\frac{424-x}{211}+\frac{424-x}{213}=0\)

\(\Leftrightarrow\)\(\left(424-x\right)\left(\frac{1}{207}+\frac{1}{209}+\frac{1}{211}+\frac{1}{213}\right)=0\)

\(\Leftrightarrow\)\(424-x=0\)   (do 1/207 + 1/209 + 1/211 + 1/213 # 0)

\(\Leftrightarrow\)\(x=424\)

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