Question

me ajudem com isso

$\frac{4}{3} x {y}^{2} - \frac{3}{8} {x}^{2} y + \frac{4}{9} {x}^{2}y - \frac{1}{4} {x}^{2} y$
determine o valor numérico para X= -1 e Y= 2​

Answer 1

Resposta:

83/36

Explicação passo-a-passo:

$\frac{4}{3} {x}^{2} y - \frac{3}{8} {x}^{2} y + \frac{4}{9} {x}^{2} y - \frac{1}{4} {x}^{2} y$

$x = - 1 \\ y = 2 \\ {x}^{2} y = {( - 1)}^{2} \times 2 = 1 \times 2 = 2$

$\frac{4}{3} \times 2 - \frac{3}{8} \times 2 + \frac{4}{9} \times 2 - \frac{1}{4} \times 2 = \\ \\ \frac{8}{3} - \frac{6}{8} + \frac{8}{9} - \frac{2}{4} = \\ \\ \frac{24}{9} - \frac{3}{4} + \frac{8}{9} - \frac{2}{4} = \\ \\ \frac{32}{9} - \frac{5}{4} = \\ \\ \frac{(32 \times 4) - (5 \times 9)}{9 \times 4} = \\ \\ \frac{128 - 45}{36} = \\ \\ \frac{83}{36}$