Question

14. Rezolvaţi în R inecuația
sa fie cu acea și coloniță ecuațiilor așa ne învață la școala ​

Answer 1

Răspuns:

$a)\:(x-1)^2+7 > (x+4)^2\\\not x^2-2x+1+7 > \not x^2+8x+16\\-2x+1+7 > 8x+16\\-2x+8 > 8x+16\\-2x-8x > 16-8\\-10x > 8\\x < -\frac{8}{10}^{(2}\\\bf\red{x < -\frac{4}{5}}$

$b)\:(1+x)^2+3x^2 < (2x-1)^2+7\\\not1+2x+x^2+3x^2 < 4x^2-4x+\not1+7\\2x+\not4x^2 < \not4x^2-4x+7\\2x < -4x+7\\6x < 7\\\bf\red{x < \frac{7}{6}}$

$c)\:(x+3)(x-2)\geq(x+2)(x-3)\\\not x^2-2x+3x-\not6\geq \not x^2-3x+2x-\not6\\-2x+3x\geq-3x+2x\\x\geq-x\\x+x\geq0\\2x\geq0\\\bf\red{x\geq0}$

$d)\:(x+1)(x-4)+4 > (x+2)(x-3)-x\\\not x^2-4x+x-4+4 > \not x^2-3x+2x-6-x\\-\not3x > -\not3x+2x-6-x\\0 > x-6\\-x > -6\\\bf\red{x < 6}$

$e)\:2(x-3)-11\leq(x-5)(x+5)-(x-1)^2\\\not2x-6-11\leq \not x^2-25-\not x^2+\not2x-1\\-6-11\leq-25-1\\-17\leq-26\:\:\:\implies\:\:\bf\red{\O}$

$f)\:(x-3)(x+3)-(x-2)^2 < 5(x+\frac{4}{5})\\\not x^2-9-\not x^2+4x-4 < 5x+4\\-13+4x < 5x+4\\4x-5x < 4+13\\-x < 17\\\bf\red{x > -17}$