Lời giải:
PT \(\Leftrightarrow \frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=4\)
\(\Leftrightarrow \frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=4\)
\(\Leftrightarrow (416-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=4\)
\(\Rightarrow 416-x=\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
\(\Rightarrow x=416-\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
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