Lời giải:
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=4\)
\(\Leftrightarrow \frac{315-x}{101}-1+\frac{313-x}{103}-1+\frac{311-x}{105}-1+\frac{309-x}{107}-1=0\)
\(\Leftrightarrow \frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow (416-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra $416-x=0$
\(\Rightarrow x=416\)
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