Question

Resuelve los cuadrados
$(2p-3q-4r)^{2}$
$(\frac{1}{2}a+\frac{2}{3}b+\frac{3}{5})^{2}$

Answer 1

Respuesta:

$(2p-3q-4r)^{2}\\= (2p-3q-4r)(2p-3q-4r)\\= 4p^{2}-6pq-8pr-6pq+9q^{2}+12qr-8pr+12qr+16r^{2}\\=4p^{2}+9q^{2}+16r^{2} -12pq -16pr+24qr$

$(\frac{a}{2}+\frac{2b}{3}+\frac{3}{5})^{2} \\=(\frac{a}{2}+\frac{2b}{3}+\frac{3}{5})(\frac{a}{2}+\frac{2b}{3}+\frac{3}{5})\\=\frac{a^{2}}{4} +\frac{ab}{3} + \frac{3a}{10}+\frac{ab}{3} + \frac{4b^{2}}{9} +\frac{2b}{5} + \frac{3a}{10} + \frac{2b}{5} + \frac{9}{25} \\=\frac{a^{2}}{4}+\frac{4b^{2}}{9}+\frac{9}{25}+\frac{3a}{5} + \frac{2ab}{3} + \frac{4b}{5}$

Si te pide poner todo bajo un mismo denominador me avisas.