Question

ДОПОМОЖІТЬ БУДЬ ЛАСКА! Розв'язати нерівність : (x+3)²-16 ≥ (1-2x)²;​

Answer 1

$\displaystyle\bf\\(x+3)^{2} -16\geq (1-2x)^{2} \\\\x^{2} +6x+9-16\geq 1-4x+4x^{2} \\\\x^{2} +6x+9-16- 1+4x-4x^{2} \geq 0\\\\-3x^{2} +10x-8\geq 0\\\\3x^{2} -10x+8\leq 0\\\\3x^{2} -10x+8=0\\\\D=(-10)^{2} -4\cdot 3\cdot 8=100-96=4=2^{2} \\\\\\x_{1} =\frac{10-2}{6}=\frac{8}{6} =1\frac{1}{3} \\\\\\x_{2} =\frac{10+2}{6} =2\\\\\\3x^{2} -10x+8=3\Big(x-1\frac{1}{3} \Big)\Big(x-2\Big)\\\\\\3\Big(x-1\frac{1}{3} \Big)\Big(x-2\Big)\leq 0\\\\\\\Big(x-1\frac{1}{3} \Big)\Big(x-2\Big)\leq 0$

$\displaystyle\bf\\+ + + + + \Big[1\frac{1}{3} \Big] - - - - - \Big[2\Big] + + + + + \\\\\\Otvet \ : \ x\in\Big[1\frac{1}{3} \ ; \ 2\Big]$