используем два выражения
<span>a³+b³=(a+b)(a²-ab+b²)</span>
<span /><span>(a-b)²=a²-2ab+b²</span>
<span /><span>(67^3+52^3) / 119-67 * 52 =(67+52)(67^2-67*52+52^2)/119 - 67*52= 119*(67^2-67*52+ 52^2)/119-67*52= 67^2-2*67*52+52^2=(67-52)^2=15^2=225
</span>
2\6 это \24,значит \24 *\24= 24
![\frac{ x^{2} - y^{2} }{ x^{2} +2xy+ y^{2} }= \frac{(x-y)(x+y)}{(x+y) ^{2} }= \frac{x-y}{x+y}\\\\ \frac{23,5-6,5}{23,5+6,5} = \frac{17}{30}\\\\\\ \frac{2x}{4x+3} \geq \frac{1}{2} \\\\ \frac{2x}{4x+3}- \frac{1}{2} \geq 0\\\\ \frac{4x-4x-3}{2(4x+3)} \geq 0\\\\ -\frac{3}{2(4x+3)} \geq 0\\\\ \frac{1}{4x+3} \leq 0\\\\4x+3 \leq 0\\\\4x \leq -3\\\\x \leq -0,75](https://tex.z-dn.net/?f=+%5Cfrac%7B+x%5E%7B2%7D+-+y%5E%7B2%7D+%7D%7B+x%5E%7B2%7D+%2B2xy%2B+y%5E%7B2%7D+%7D%3D+%5Cfrac%7B%28x-y%29%28x%2By%29%7D%7B%28x%2By%29+%5E%7B2%7D+%7D%3D+%5Cfrac%7Bx-y%7D%7Bx%2By%7D%5C%5C%5C%5C+%5Cfrac%7B23%2C5-6%2C5%7D%7B23%2C5%2B6%2C5%7D+%3D++%5Cfrac%7B17%7D%7B30%7D%5C%5C%5C%5C%5C%5C+%5Cfrac%7B2x%7D%7B4x%2B3%7D+%5Cgeq++%5Cfrac%7B1%7D%7B2%7D+%5C%5C%5C%5C+%5Cfrac%7B2x%7D%7B4x%2B3%7D-+%5Cfrac%7B1%7D%7B2%7D+%5Cgeq+0%5C%5C%5C%5C+%5Cfrac%7B4x-4x-3%7D%7B2%284x%2B3%29%7D+%5Cgeq+0%5C%5C%5C%5C+-%5Cfrac%7B3%7D%7B2%284x%2B3%29%7D++%5Cgeq+0%5C%5C%5C%5C+%5Cfrac%7B1%7D%7B4x%2B3%7D+%5Cleq+0%5C%5C%5C%5C4x%2B3+%5Cleq+0%5C%5C%5C%5C4x+%5Cleq+-3%5C%5C%5C%5Cx+%5Cleq+-0%2C75++++++++++)
ОДЗ: 4x + 3 ≠ 0 ⇒ x ≠ - 0,75
Ответ: (- ∞ ; - 0,75)
Так как обе части данного уравнения неотрицательны, то возведём обе части в квадрат →
![\sqrt{3x + 27} = 6 \\ 3x + 27 = {6}^{2 } \\ 3x + 27 = 36 \\ 3x = 36 - 27 \\ 3x = 9 \\ x = \frac{9}{3} = 3 \\](https://tex.z-dn.net/?f=+%5Csqrt%7B3x+%2B+27%7D+%3D+6+%5C%5C+3x+%2B+27+%3D+%7B6%7D%5E%7B2+%7D+%5C%5C+3x+%2B+27+%3D+36+%5C%5C+3x+%3D+36+-+27+%5C%5C+3x+%3D+9+%5C%5C+x+%3D+%5Cfrac%7B9%7D%7B3%7D+%3D+3+%5C%5C+)
ПРОВЕРКА:
![\sqrt{3 x + 27} = \sqrt{3 \times 3 + 27} = \sqrt{9 + 27} = \sqrt{36} = 6](https://tex.z-dn.net/?f=+%5Csqrt%7B3+x+%2B+27%7D+%3D+%5Csqrt%7B3+%5Ctimes+3+%2B+27%7D+%3D+%5Csqrt%7B9+%2B+27%7D+%3D+%5Csqrt%7B36%7D+%3D+6)
верно
ОТВЕТ: 3