Nhận xét:
$|x+\frac{1}{1.5}|+|x+\frac{1}{5.9}|+...+|x+\frac{1}{397.401}|$$\geq0∀x$
$⇒101x≥0⇔x≥0$
$⇒|x+\frac{1}{1.5}|+|x+\frac{1}{5.9}|+...+|x+\frac{1}{397.401}|$
$=(x+\frac{1}{1.5})+(x+\frac{1}{5.9})+...+(x+\frac{1}{397.401})=A$
Số cặp ngoặc đơn là: $\frac{397-1}4+1=100$ (cặp)
$⇒A=100x+\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{393.397}+\frac{4}{397.401}=101x$
$⇔x=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{393.397}+\frac{4}{397.401}$
$⇔4x=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{393.397}+\frac{4}{397.401}$
$⇔4x=1-\frac{1}5+\frac{1}5-\frac{1}9+...+\frac{1}{393}-\frac{1}{397}+\frac{1}{397}-\frac{1}{401}$
$⇔4x=1-\frac{1}{401}$
$⇔4x=\frac{400}{401}$
$⇔x=\frac{400}{401}÷4=\frac{100}{401}$
