Question

me ajudem por favor!!!​

Answer 1

a) 2 x² + 5 x . 1 =0

$2x^2+5x\cdot \:1=0\quad :\quad x=0,\:x=-\frac{5}{2}\\\\2x^2+5x\cdot \:1:\quad 2x^2+5x\\\\2x^2+5x=0\\\\\\\mathrm{Para\:uma\:equacao\:de\:segundo\:grau\:da\:forma\:}ax^2+bx+c=0\mathrm{\:as\:solucoes\:são\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-5+\sqrt{5^2-4\cdot \:2\cdot \:0}}{2\cdot \:2}:\quad 0\\\\x=\frac{-5-\sqrt{5^2-4\cdot \:2\cdot \:0}}{2\cdot \:2}:\quad -\frac{5}{2}\\\\\\x=0,\:x=-\frac{5}{2}$

b)x² . 16=0

$x^2\cdot \:16=0\quad :\quad x=0\\\\x^2\cdot \:16=0\\\\\frac{x^2\cdot \:16}{16}=\frac{0}{16}\\\\x^2=0\\\\x=0$

c)-4x³=0

$-4x^3=0\quad :\quad x=0\\\\-4x^3=0$

$\frac{-4x^3}{-4}=\frac{0}{-4}$

$x^3=0\\\\\mathrm{Aplicar\:a\:regra}\:x^n=0\quad \Rightarrow \quad \:x=0\\\\x=0$

d)x² .1=0

$x^2\cdot \:1=0\quad :\quad x=0\\\\x^2\cdot \:1=0\\\\x^2\cdot \:1=x^2\\x^n=0\quad \Rightarrow \quad \:x=0$

e)5x² . 7x =0

$5x^{2.7}x=0\quad :\quad x=0\\\\5x^{2.7}x=0\\\\\mathrm{Usando\:o\:principio\:do\:fator\:zero:\quad \:Se}\:ab=0 entao \:a=0\:\mathrm{ou}\:b=0\:\left(\mathrm{ou\:ambos}\:a=0\:\mathrm{e}\:b=0\right)\\\\\\x^{2.7}=0:\quad x=0\\\\x=0\\x=0\\\mathrm{Verifique\:solucoes}:\quad x=0\space\mathrm{Verdadeiro}\\\\\\\\x=0$