Question

Допоможіть будь ласка даю(15)​

Answer 1

Ответ:

$\displaystyle \boldsymbol {x\leq -\frac{11}{19} }$

Объяснение:

$\displaystyle \left \{ {{ \displaystyle\frac{5x+1}{3} \geq 4(2x+1)\hfill \atop {(x+2)(x-5)\geq x(x+1)-10}} \right. \left \{ {{5x+1\geq 12(2x+1)\hfill} \atop {x^2-5x+2x-10\geq x^2+x-10}} \right. \\\\\\\left \{ {{-11\geq 19x\hfill} \atop {x^2-5x+2x-10\geq x^2+x-10}} \right. \left \{ {{ \displaystyle x\leq -\frac{11}{19} } \atop {-4x\geq 0}} \right. \left \{ {{ \displaystyle x\leq -\frac{11}{19} } \atop {x\leq 0}} \right. \Rightarrow\; \displaystyle x\leq -\frac{11}{19}$