Rozwiąż nierówność;
a) [tex]2-\frac{x+3}{3} \ \textless \ \frac{2x-3}{2}[/tex]
b) [tex]4(x-3)[/tex]²[tex]- (2x-5)[/tex]²[tex]\geq 2[/tex]
a) [tex]2-\frac{x+3}{3} \ \textless \ \frac{2x-3}{2}[/tex]
b) [tex]4(x-3)[/tex]²[tex]- (2x-5)[/tex]²[tex]\geq 2[/tex]
Odpowiedź:
a) x > 1 7/8
b) x ≤ 2 1/4
Szczegółowe wyjaśnienie:
[tex]2-\frac{x+3}{3} < \frac{2x-3}{2} |\cdot6\\12-2(x+3) < 3(2x-3)\\12-2x-6 < 6x-9\\-2x+12-6 < 6x-9\\-2x+6 < 6x-9\\-2x-6x < -9-6\\-8x < -15|:(-8)\\x > 1\frac{7}{8}[/tex]
[tex]4(x-3)^2-(2x-5)^2\geq 2\\4(x^2-6x+9)-(4x^2-20x+25)\geq 2\\4x^2-24x+36-4x^2+20x-25\geq 2\\-24x+20x+36-25\geq 2\\-4x+11\geq 2\\-4x\geq 2-11\\-4x\geq -9 |:(-4)\\x\leq 2\frac{1}{4}[/tex]