Với mọi x ta có :
+) \(\left|x+\dfrac{1}{1.3}\right|\ge0; \)
+) \(\left|x+\dfrac{1}{3.5}\right|\ge0;\)
.....................................
+) \(\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Leftrightarrow\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+.......+\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Leftrightarrow50x\ge0\)
\(\Leftrightarrow x\ge0\)
Khi \(x\ge0\) ta được :
+) \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3}\)
+) \(\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5}\)
.............................................
+) \(\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\)
\(\Leftrightarrow\left(x+\dfrac{1}{1.3}\right)+\left(x+\dfrac{1}{3.5}\right)+......+\left(x+\dfrac{1}{97.99}\right)=50x\)
\(\Leftrightarrow49x+\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{97.99}\right)=50x\)
\(\Leftrightarrow x=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Leftrightarrow x=\dfrac{16}{99}\)
Vậy...
