Question

(9×-4)²=81×-72×+16 ayuda q-q porfavor ​

Answer 1

Respuesta:

$x=1,\:x=0$

Explicación paso a paso:

$\left(9x-4\right)^2=81x-72x+16$

$81x^2-72x+16=9x+16$

$81x^2-72x+16-16=9x+16-16$

$81x^2-72x=9x$

$81x^2-72x-9x=9x-9x$

$81x^2-81x=0$

$x_{1,\:2}=\frac{-\left(-81\right)\pm \sqrt{\left(-81\right)^2-4\cdot \:81\cdot \:0}}{2\cdot \:81}$

$x_{1,\:2}=\frac{-\left(-81\right)\pm \:81}{2\cdot \:81}$

$x_1=\frac{-\left(-81\right)+81}{2\cdot \:81},\:x_2=\frac{-\left(-81\right)-81}{2\cdot \:81}$

$x_=\frac{-\left(-81\right)+81}{2\cdot \:81}=1$

$x_=\frac{-\left(-81\right)-81}{2\cdot \:81}=0$

$x=1,\:x=0$

(Espero que te haya servido de ayuda)