Question

NECESITO QUE ALGUIEN ME AYUDE CON ESTE EJERCICIO, PORFA

Answer 1

$\left(\frac{1}{4}-1\right)^{-2}+\frac{2-\left(-3\right)\cdot \:5+1}{9}$
$\left(\frac{1}{4}-1\right)^{-2}=\left(-\frac{3}{4}\right)^{-2}=\frac{1}{\left(-\frac{3}{4}\right)^2}=\frac{1}{\left(\frac{3}{4}\right)^2}=\frac{1}{\frac{3^2}{4^2}}=\frac{4^2}{3^2}$
$\frac{2-\left(-3\right)\cdot \:5+1}{9}=\frac{2+3\cdot \:5+1}{9}=\frac{18}{9}=2$
$=\frac{4^2}{3^2}+2=\frac{16}{9}+2=\frac{2}{1}+\frac{16}{9}=\frac{2\cdot \:9}{9}+\frac{16}{9}$
$=\frac{2\cdot \:9+16}{9}=\frac{34}{9}$

Answer 2

(1/4 -1)⁻²+2-(-3)×5+1/9
(1- 4/4)⁻² +2+3×6/9
(3/4)⁻² + 6×6/9
aqui se invierte la fraccion para volver el exponente positivo (4/3)²+36/9
16/9+4
16+4/9
52/9