Odpowiedź:
[tex]\huge\boxed{a) \ x\in\langle-\frac{2}{3};+\infty)}[/tex]
[tex]\huge\boxed{b) \ x\in(-\infty;4\rangle\cup\langle6;+\infty)}[/tex]
Szczegółowe wyjaśnienie:
Podpunkt a)
[tex](x-2)^2+x\leq x^2+6\\\\x^2-2\cdot x\cdot2+2^2+x\leq x^2+6\\\\x^2-4x+4+x\leq x^2+6\\\\x^2-4x+x-x^2\leq6-4\\\\-3x\leq2 \ \ /:(-3)\\\\x\geq-\frac{2}{3}\\\\x\in\langle-\frac{2}{3};+\infty)[/tex]
Podpunkt b)
[tex]x^2+24\geq10x\\\\x^2-10x+24\geq0\\\\a=1, \ b=-10, \ c=24\\\\\Delta=b^2-4ac\Rightarrow(-10)^2-4\cdot1\cdot24=100-96=4\\\\\sqrt{\Delta}=\sqrt4=2\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}\Rightarrow\frac{-(-10)-2}{2\cdot1}=\frac{10-2}{2}=\frac{8}{2}=4\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\Rightarrow\frac{-(-10)+2}{2\cdot1}=\frac{10+2}{2}=\frac{12}{2}=6\\\\x\in(-\infty;4\rangle\cup\langle6;+\infty)[/tex]
a > 0, ramiona paraboli są skierowane do góry
