Rozwiąż nierówność.
a) x²-(x+3) (x-3) ≤ 6x
b) (4-x) (2x+3)+2x² < 6
c) (2x-1) (3x-1) - (3x-2) (2x-3) ≥ 0
d) (4-6x) (2x + 1) + (4x-5) (3x - 1) > x


Odpowiedź :

[tex]a)\\\\ x^2-(x+3) (x-3) \leq 6x\\\\x^2-(x^2-9)\leq 6x\\\\x^2-x^2+ 9-6x\leq 0\\\\-6x\leq -9\ \ |:(-6)\\\\x\geq \frac{9}{6}\\\\x\geq \frac{3}{2}[/tex]

[tex]b )\\\\ (4-x) (2x+3)+2x^2 < 6 \\\\8x+12-2x^2-3x+2x^2<6\\\\5x<6-12\\\\5x<-6\ \ |:5\\\\x<-\frac{6}{5}[/tex]

[tex]c)\\\\(2x-1) (3x-1) - (3x-2) (2x-3) \geq 0 \\\\6x^2-2x-3x+1-(6x^2-9x-4x+6)\geq 0\\\\6x^2-5x +1- 6x^2+9x+4x-6 \geq 0\\\\ 8x\geq 5\ \ |:8\\\\x\geq \frac{5}{8}[/tex]

[tex]d)\\\\ (4-6x) (2x + 1) + (4x-5) (3x - 1) > x\\\\8x+4-12x^2-6x+12x^2-4x-15x+5>x\\\\ -17x+9>x\\\\-17x-x>-9\\\\-18x>-9\ \ |:(-18)\\\\x<\frac{9}{18}\\\\x<\frac{1}{2}[/tex]