Rozwiąż nierówność przykłady na zdjęciu

[tex] {(3x - 1)}^{2} \leqslant {(5x - 3)}^{2} \\ 9 {x}^{2} - 6x + 1 \leqslant 25 {x}^{2} - 30x + 9 \\ 16 {x}^{2} - 24x + 8 \geqslant 0 \\ 2 {x}^{2} - 3x + 1 \geqslant 0 \\ delta \\ {( - 3)}^{2} - 4 \times 2 \times 1 = 9 - 8 = 1 \\ x = \frac{3 - 1}{4} = \frac{1}{2} \\ x = \frac{3 + 1}{4} = 1 \\ x = ( - \infty . \frac{1}{2} > i < 1. \infty )[/tex]

[tex] {(3x - 1)}^{2} - {(4x - 2)}^{2} \leqslant 0 \\ 9 {x}^{2} - 6x + 1 - 16 {x}^{2} + 16x - 4 \leqslant 0 \\ - 7 {x}^{2} - 10x - 3 \leqslant 0 \\ 7 {x}^{2} + 10x + 3 \geqslant 0 \\ delta \\ {10}^{2} - 4 \times 7 \times 3 = 100 - 84 = 16 \\ x = \frac{ - 10 - 4}{14} = - 1 \\ x = - \frac{ - 10 + 4}{14} = - \frac{ - 6}{14} = - \frac{3}{7} \\ odp \\ x = ( - \infty . \frac{ - 3}{7} > i < - 1. \infty )[/tex]

Magdaviz Magdaviz · Answer 2

Odpowiedź:

[tex]a)\\\\(3x-1)^2\leq (5x-3)^2\\\\9x^2-6x+1\leq 25x^2-30x+9\\\\9x^2-6x+1-25x^2+30x-9\leq 0\\\\-16x^2+24x-8\leq 0\ \ |:(-8)\\\\2x^2-3x+1\geq 0\\\\a=2\ \ ,\ \ b=-3\ \ ,\ \ c=1\\\\\Delta=b^2-4ac\\\\\Delta=(-3)^2-4\cdot2\cdot1=9-8=1\\\\\sqrt{\Delta}=\sqrt{1}=1\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2\cdot2}=\frac{3-1}{4}=\frac{2}{4}=\frac{1}{2}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2\cdot2}=\frac{3+1}{4}=\frac{4}{4}=1[/tex]

[tex]b)\\\\(3x-1)^2-(4x-2)^2\leq 0\\\\9x^2-6x+1-(16x^2-16x+4)\leq 0\\\\9x^2-6x+1-16x^2+16x-4\leq 0\\\\-7x^2+10x-3\leq 0\ \ |\cdot(-1)\\\\7x^2-10x+3\geq 0\\\\a=7\ \ ,\ \ b=-10\ \ ,\ \ c=3\\\\\Delta=b^2-4ac\\\\\Delta=(-10)^2-4\cdot7\cdot3=100-84=16\\\\\sqrt{\Delta}=\sqrt{16}=4\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4}{2\cdot7}=\frac{10-4}{14}=\frac{6}{14}=\frac{3}{7}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4}{2\cdot7}=\frac{10+4}{14}=\frac{14}{14}=1[/tex]