\(P=\left|x-\frac{20}{11}\right|+\frac{2}{3}\ge\frac{2}{3}\)( vì \(\left|x-\frac{20}{11}\right|\ge0\forall x\))
Min P = 2/3
\(\Leftrightarrow\left|x-\frac{20}{11}\right|=0\)
\(\Leftrightarrow x=\frac{20}{11}\)
P = | x - 20/11 | + 2/3
| x - 20/11 | ≥ 0 ∀ x => | x - 20/11 | + 2/3 ≥ 2/3
Đẳng thức xảy ra <=> x - 20/11 = 0 => x = 20/11
=> MinP = 2/3 <=> x = 20/11