podwójna nierówność

[BarT123oks]

\(|x^2-5x+6|\leq |x-3| < x^2-|3x-x^2|\)

[eresh]

\(|x^2-5x+6|\leq |x-3| < x^2-|3x-x^2|\)

\(|(x-3)(x-2)|\leq |x-3|\;\;\;\wedge\;\;\;|x-3|<x^2-|x(x-3)|\\
\)

\(|(x-3)(x-2)|\leq |x-3|
\)
1. \(x\in (-\infty, -2)\)
\((x-3)(x-2)\leq -(x-3)\\
(x-3)(x-2)+(x-3)\leq 0\\
(x-3)(x-1)\leq 0\\
x\in [1,2)\)

2. \(x\in [-2,3)\)
\(-(x-3)(x-2)\leq -(x-3)\\
(x-3)(x-2)\geq (x-3)\\
(x-3)(x-2-1)\geq 0\\
(x-3)(x-3)\geq 0\\
x\in [-2,3)\)

3. \(x\in [3,\infty)\)
\((x-3)(x-2)\leq (x-3)\\
(x-3)(x-3)\leq 0\\
x=3\)

\(x\in [1,3]\)


\(|x-3|<x^2-|x(x-3)|\\
\)
1. \(x\in (-\infty, 0)\)
\(-(x-3)<x^2-x(x-3)\\
-x+3<x^2-x^2+3x\\
-4x<-3\\
x>\frac{3}{4}\\
\emptyset
\)

2. \(x\in [0,3)\)
\(-x+3<x^2+x^2-3x\\
2x^2-2x-3>0\\
x\in (\frac{1+\sqrt{7}}{2},3)\)

3. \(x\in [3,\infty)\)
\(x-3<x^2-x(x-3)\\
x-3<x^2-x^2+3x\\
-2x<3\\
x>-\frac{3}{2}\\
x\in [3,\infty)\)

\(x\in (\frac{1+\sqrt{7}}{2},\infty)\)

Odpowiedź: \(x\in (\frac{1+\sqrt{7}}{2},3]\)