`C = |x-1|+|2x-3|`
Ta có: `|x-1|≥0 ; |2x-3|≥1`
`=> C = |x-1|+|2x-3| ≥ 1`
Dấu "=" có khi:
`C = |x-1|+|2x-3| = 1`
⇔ $\left \{ {{|x-1|=0} \atop {|2x-3|=1}} \right.$
⇔ $\left \{ {{x-1=0} \atop {\left[ \begin{array}{l}2x-3=1\\2x-3=-1\end{array} \right. }} \right.$
⇔$\left \{ {{x=1} \atop {\left[ \begin{array}{l}2x=4\\2x=2\end{array} \right. }} \right.$
⇔$\left \{ {{x=1} \atop {\left[ \begin{array}{l}x=2\\x=1\end{array} \right. }} \right.$
`⇔ x = 1`
---------------------------------
`D = x^2-4x+5`
`= x^2-4x1/2+(1/2)^2+19/4`
`= x^2-4x1/2+(1/2)^2 +19/4`
`= x^2-2x+(1/2)^2+ 19/4`
`= (x - 1/2)^2 + 19/4`
Ta có: `(x - 1/2)^2 ≥ 0`
`=> (x - 1/2)^2 + 19/4 ≥ 19/4`
`=>` Dấu "=" xảy ra khi:
`(x - 1/2)^2 + 19/4 = 19/4`
`=> (x - 1/2)^2 = 0`
`=> x - 1/2 = 0`
`=> x = 1/2`
(Chúc bạn học tốt)
