Question

Alguém me ajuda, pelo amor deu DEUS! tem que chegar a ZERO o resultado!!! ​

Answer 1

Explicação passo-a-passo:

$[\frac{3(\frac{-3}{4})^{-2}+6(\frac{3^{-1}}{4})-4}{7(\frac{-3}{4})^{-1}+2}]^{-1}+4=$

$\frac{7(\frac{-3}{4})^{-1}+2}{3(\frac{-3}{4})^{-2}+6(\frac{3^{-1}}{4})-4}+4=$

$\frac{7(\frac{4}{-3})+2}{3(\frac{4}{-3})^{2}+6(\frac{\frac{1}{3}}{4})-4}+4=$

$\frac{7(\frac{4}{-3})+2}{3(\frac{16}{9})+6(\frac{\frac{1}{3}}{4})-4}+4=$

$\frac{7(\frac{4}{-3})+2}{3(\frac{16}{9})+6(\frac{1}{12})-4}+4=$

$\frac{\frac{28}{-3}+2}{\frac{48}{9}+\frac{6}{12}-4}+4=$

$\frac{-\frac{28}{3}+\frac{6}{3}}{\frac{16}{3}+\frac{1}{2}-4}+4=$

$\frac{-\frac{22}{3}}{\frac{32}{6}+\frac{3}{6}-4}+4=$

$\frac{-\frac{22}{3}}{\frac{35}{6}-4}+4=$

$\frac{-\frac{22}{3}}{\frac{35}{6}-\frac{24}{6}}+4=$

$\frac{-\frac{22}{3}}{\frac{11}{6}}+4=$

$-\frac{22}{3}\times\frac{6}{11}+4=$

$-\frac{2}{1}\times\frac{2}{1}+4=$

$-\frac{4}{1}+4=$

$-4+4=$

$0$