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Rozwiąż nierówności:

a) [tex]x^{5} + 2 x^{4} - 5 x^{3} - 10 x^{2} \geq 0[/tex]

b) [tex]-2x^3-6x^2-4x\ \textless \ 0[/tex]

c) [tex]-6x^3+13x^2-4x-3\geq 0[/tex].


Odpowiedź :

[tex]a)\\\\x^5+2x^4-5x^3-10x^2 \geq 0\\x^4(x+2)-5x^2(x+2) \geq 0\\(x^4-5x^2)(x+2) \geq 0\\x^2(x^2-5)(x+2) \geq 0\\\\x^2 \geq 0\\x \geq 0 \text{ v } x \leq 0 \to x \in R[/tex]

[tex]x^2-5 \geq 0\\(x-\sqrt5)(x+\sqrt5) > 0\\x-\sqrt5 > 0 \\x > \sqrt5\\\\x+\sqrt5 > 0\\x > -\sqrt5[/tex]

[tex]x+2 \geq 0\\x \geq -2[/tex]

[tex]\bold{x\in < -\sqrt5; -2 > U \{0\} U < \sqrt5; \infty)}[/tex]

[tex]b)\\\\-2x^3-6x^2-4x < 0\\-2x(x^2+3x+2) < 0\\\Delta=3^2-4*1*2=9-8=1\\x_1=\frac{-3-1}2=-2\\x_2=\frac{-3+1}2=-1\\-2x(x+2)(x+1) < 0\\\\-2x < 0 /:(-2)\\x > 0\\\\-(x+2) < 0\\-x-2 < 0\\-x < 2\\x > -2\\\\x+1 < 0\\x < -1\\\\\bold{x\in(-2; -1)U(0; \infty)}[/tex]

[tex]c)\\\\-6x^3+13x^2-4x-3 \geq 0\\-(6x^3-13x^2+4x+3)\geq 0\\-(x-1)(6x^2-7x-3) \geq 0\\\Delta=(-7)^2-4*6*(-3)=49+72=121\\\sqrt{\Delta}=11\\x_1=\frac{7-11}{12}=\frac{-4}{12}=-\frac13\\x_2=\frac{7+11}{12}=\frac{18}{12}=\frac32\\-(x-1)(x+\frac13)(x-\frac32) \geq 0\\-(x-1)(3x+1)(2x-3)\geq 0\\\\\\x-1 \geq 0\\x \geq 1\\\\-(3x+1) \geq 0\\-3x-1 \geq 0\\-3x \geq 1 /:(-3)\\x \leq -\frac13\\\\[/tex]

[tex]-(2x-3) \geq 0\\-2x+3 \geq 0\\-2x \geq -3 /:(-2)\\x \leq \frac32\\\\\bold{x\in (-\infty; -\frac13 > U < 1; \frac32 > }[/tex]