Rozwiązane

rozwiąż nierówności
(2x+1)2+(x-3)2 < 10
3x – (1 – x)2 > x2 - 4

Odpowiedź :

Kontonawszystkoviz

[tex](2x + 1 {)}^{2} + (x - 3 {)}^{2} < 10 \\ 4 {x}^{2} + 4x + 1 + {x}^{2} - 6x + 9 < 10 \\ 5 {x}^{2} - 2x + 10 < 10 \\ 5 {x}^{2} - 2x < 0 \\ x(5x - 2) < 0 \\ x = 0 \: \: \: \: 5x - 2 = 0 \\ \: \: \: \: \: \: \: \: 5x = 2 \\ \: \: \: \: \: \: \: \: \: x = \frac{2}{5} [/tex]

załącznik 1 niebieski

[tex]3x - (1 - x {)}^{2} > {x}^{2} - 4 \\ 3x - (1 - 2x + {x}^{2} ) > {x}^{2} - 4 \\ 3x - 1 + 2x - {x}^{2} > {x}^{2} - 4 \\ - {x}^{2} - {x}^{2} + 3x + 2x - 1 + 4 > 0 \\ - 2 {x}^{2} + 5x + 3 > 0 \\ ∆ = {5}^{2} - 4 \times ( - 2) \times 3 = 25 + 24 = 49 \\ \sqrt{∆} = 7 \\ x_{1} = \frac{ - 5 - 7}{2 \times ( - 2)} = \frac{ - 12}{ - 4} = 3 \\ x _{2} = \frac{ - 5 + 7}{ - 4} = \frac{2}{ - 4} = - \frac{1}{2} [/tex]

załącznik 2 czerwony

Zobacz obrazek Kontonawszystko
Zobacz obrazek Kontonawszystko
Basetlaviz

1.

[tex](2x+1)^{2}+(x-3)^{2} < 10\\\\4x^{2}+4x+1 + x^{2}-6x+9 -10< 0\\\\5x^{2}-2x< 0\\\\M. \ zerowe:\\\\5x^{2}-2x = 0\\\\x(5x-2) = 0\\\\x = 0 \ \vee \ 5x -2 = 0\\\\x = 0 \ \vee \ 5x = 2 \ \ /:5\\\\x = 0 \ \vee \ x = \frac{2}{5}\\\\a > 0, to \ ramiona \ paraboli \ skierowane \ w \ gore, \ wartosci < 0 \ znajduja \ sie \ pod \ osia \ OX\\\\x \in \ (0; \frac{2}{5})[/tex]

2.

[tex]3x - (1-x)^{2}>x^{2}-4\\\\3x - (1^{2}-2x+x^{2})>x^{2}-4\\\\3x-1+2x-x^{2}-x^{2}+4 > 0\\\\-2x^{2}+5x+3 > 0\\\\\Delta = 5^{2}-4\cdot(-2)\cdot3 = 25+24 = 49\\\\\sqrt{\Delta} = \sqrt{49} = 7\\\\x_1 = \frac{-5+7}{-4} = \frac{2}{-4} = -\frac{1}{2}\\\\x_2 = \frac{-5-7}{-4} = \frac{-12}{-4} = 3\\\\a < 0, \ ramiona \ paraboli \ skierowane \ do \ dolu, \ wartosci > 0 \ znajduja \ sie \ nad \ osia \ OX\\\\x \in (-\frac{1}{2}; 3)[/tex]