(0,5 điểm) Tìm x , biết: ∣ ∣ x + 1 101 ∣ ∣ + ∣ ∣ x + 2 101 ∣ ∣ + ∣ ∣ x + 3 101 ∣ ∣ + . . . + ∣ ∣ x + 100 101 ∣ ∣ = 101 x .

Hướng dẫn giải

Nhận thấy \(\left| {x + \frac{1}{{101}}} \right| \ge 0;\) \(\left| {x + \frac{2}{{101}}} \right| \ge 0;\) \(\left| {x + \frac{3}{{101}}} \right| \ge 0\)…..;\(\left| {x + \frac{{100}}{{101}}} \right| \ge 0\)

Do đó, \(\left| {x + \frac{1}{{101}}} \right| + \left| {x + \frac{2}{{101}}} \right| + \left| {x + \frac{3}{{101}}} \right| + ... + \left| {x + \frac{{100}}{{101}}} \right| \ge 0\).

Mà \(\left| {x + \frac{1}{{101}}} \right| + \left| {x + \frac{2}{{101}}} \right| + \left| {x + \frac{3}{{101}}} \right| + ... + \left| {x + \frac{{100}}{{101}}} \right| = 101x\) nên \(101x \ge 0\) hay \(x \ge 0\).

Với \(x \ge 0\), suy ra \(x + \frac{1}{{101}} + x + \frac{2}{{101}} + x + \frac{3}{{101}} + ... + x + \frac{{100}}{{101}} = 101x\)

\(100x + \left( {\frac{1}{{101}} + \frac{2}{{101}} + \frac{3}{{101}} + ... + \frac{{100}}{{101}}} \right) = 101x\)

\(101x - 100x = \frac{{1 + 2 + 3 + ... + 100}}{{101}}\)

\(x = \frac{{100.101}}{{2.101}}\)

\(x = 50\) (thỏa mãn)

Vậy \(x = 50\).