Question

решите неравенство (2x+5)^2≥ (5x-2)^2

Answer 1

$(2x+5)^2\geq (5x-2)^2\\\\(2x+5)^2-(5x-2)^2\geq 0\; ,\qquad A^2-B^2=(A-B)(A+B)\\\\(2x+5-5x+2)(2x+5+5x-2)\geq 0\\\\(7-3x)(7x+3)\geq 0\\\\x_1=\frac{7}{3}\; ,\; \; x_2=-\frac{3}{7}\\\\znaki:\; \; \; +++[-\frac{3}{7}\, ]---[\, \frac{7}{3}\, ]+++\\\\x\in (-\infty ,-\frac{3}{7}\, ]\cup [\, 2\frac{1}{3}\; ,+\infty )$