Question

Alguem resolve pfv!!!

Answer 1

c)

$Multplicar\ ambos\ os\ lados\ por\ 5:\\x\cdot \:5+\frac{x}{5}\cdot \:5<2x\cdot \:5-3\cdot \:5\\Simplificar:\\6x<10x-15\\\mathrm{Subtrair\:}10x\mathrm{\:de\:ambos\:os\:lados}:\\6x-10x<10x-15-10x\\Simplificar:\\-4x<-15\\Multiplicar\:ambos\:os\:lados\:por\:-1\:\left(inverte\:a\:desigualdade\right):\\\left(-4x\right)\left(-1\right)>\left(-15\right)\left(-1\right)\\Simplificar:\\4x>15\\\mathrm{Dividir\:ambos\:os\:lados\:por\:}4:\\\frac{4x}{4}>\frac{15}{4}\\Simplificar:\\x>\frac{15}{4}$

d)

$\frac{x^2}{18}-\frac{x}{6}+\frac{1}{9}>\frac{1}{9}\\\frac{x^2}{18}-\frac{x}{6}>0\\\frac{x^2}{18}\cdot \:18-\frac{x}{6}\cdot \:18>0\cdot \:18\\x^2-3x>0\\=x\left(x-3\right)\\x\left(x-3\right)>0\\x<0\quad \mathrm{ou}\quad \:x>3$

e)

$\frac{3x}{2}-\frac{1}{6}-1\ge \:10x^2-\frac{55x}{3}-\frac{10}{3}\\\frac{3x}{2}-\frac{1}{6}-1+\frac{10}{3}\ge \:10x^2-\frac{55x}{3}-\frac{10}{3}+\frac{10}{3}\\\frac{3x}{2}-\frac{1}{6}-1+\frac{10}{3}\ge \:10x^2-\frac{55x}{3}\\\frac{3x}{2}-\frac{1}{6}-1+\frac{10}{3}+\frac{55x}{3}\ge \:10x^2-\frac{55x}{3}+\frac{55x}{3}\\\frac{3x}{2}-\frac{1}{6}-1+\frac{10}{3}+\frac{55x}{3}\ge \:10x^2\\\frac{3x}{2}-\frac{1}{6}-1+\frac{10}{3}+\frac{55x}{3}-10x^2\ge \:10x^2-10x^2$

$\frac{3x}{2}-\frac{1}{6}-1+\frac{10}{3}+\frac{55x}{3}-10x^2\ge \:10x^2-10x^2\\\frac{3x}{2}\cdot \:6-\frac{1}{6}\cdot \:6-1\cdot \:6+\frac{10}{3}\cdot \:6+\frac{55x}{3}\cdot \:6-10x^2\cdot \:6\ge \:0\cdot \:6\\-60x^2+119x+13\ge \:0\\-60\left(x-\frac{119}{120}\right)^2+\frac{17281}{240}\ge \:0\\-60\left(x-\frac{119}{120}\right)^2+\frac{17281}{240}-\frac{17281}{240}\ge \:0-\frac{17281}{240}\\-60\left(x-\frac{119}{120}\right)^2\ge \:-\frac{17281}{240}$

$\left(-60\left(x-\frac{119}{120}\right)^2\right)\left(-1\right)\le \left(-\frac{17281}{240}\right)\left(-1\right)\\60\left(x-\frac{119}{120}\right)^2\le \frac{17281}{240}\\\frac{60\left(x-\frac{119}{120}\right)^2}{60}\le \frac{\frac{17281}{240}}{60}\\\left(x-\frac{119}{120}\right)^2\le \frac{17281}{14400}\\-\sqrt{\frac{17281}{14400}}\le \:x-\frac{119}{120}\le \sqrt{\frac{17281}{14400}}$

$-\sqrt{\frac{17281}{14400}}\le \:x-\frac{119}{120}\quad \mathrm{e}\quad \:x-\frac{119}{120}\le \sqrt{\frac{17281}{14400}}\\x\ge \frac{-\sqrt{17281}+119}{120}\quad \mathrm{e}\quad \:x\le \frac{\sqrt{17281}+119}{120}\\\frac{-\sqrt{17281}+119}{120}\le \:x\le \frac{\sqrt{17281}+119}{120}$