Question

Doy coronita y sigo cumplo pero me ayudan es para el lunes (⁠╥⁠﹏⁠╥⁠)


RESUELVE LO SIGUIENTE CON SU RESPECTIVA "COMPROBACIÓN"

• 9x+1=3(x² -5) - (x+2)(x+2)



• (2x - 3)² - (x+5)² = - 23​

Answer 1

Respuesta:

Ok

1) 9x+1=3(x² -5) - (x+2)(x+2)

$9x+1=3\left(x^2-5\right)-\left(x+2\right)\left(x+2\right)\\\\9x+1=3\left(x^2-5\right)-\left(x+2\right)^2\\\\9x+1=3\left(x^2-5\right)-\left(x^2+4x+4\right)\\\\9x+1=3x^2-15-\left(x^2+4x+4\right)\\\\9x+1=3x^2-15-\left(x^2+4x+4\right)\\\\9x+1=3x^2-15-x^2-4x-4\\\\2x^2-13x-20=0\\\\x_{1,\:2}=\frac{-\left(-13\right)\pm \sqrt{\left(-13\right)^2-4\cdot \:2\left(-20\right)}}{2\cdot \:2}\\\\x_{1,\:2}=\frac{-\left(-13\right)\pm \sqrt{329}}{2\cdot \:2}\\x=\frac{13+\sqrt{329}}{4} , x=\frac{13-\sqrt{329}}{4}$

2) (2x - 3)² - (x+5)² = - 23​

$\left(2x-3\right)^2-\left(x+5\right)^2=-23\\\\3x^2-22x-16=-23\\\\3x^2-22x-16+23=-23+23\\\\3x^2-22x+7=0\\\\x_{1,\:2}=\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\cdot \:3\cdot \:7}}{2\cdot \:3}\\\\x=7 , \ x=1/3$