Question

5. Resuelve los productos ayudaaaaaaa :-(​

Answer 1

Respuesta:

5. Resuelve los productos ayudaaaaaaa :-(​

Explicación paso a paso:

$\left(x^m-y^n\right)^2\\=\left(x^m\right)^2-2x^my^n+\left(y^n\right)^2\\=x^{2m}-2x^my^n+y^{2n}$

$\left(a^{m+3}+b^{n+2}\right)^2\\=\left(a^{m+3}\right)^2+2a^{m+3}b^{n+2}+\left(b^{n+2}\right)^2\\=a^{2\left(m+3\right)}+2a^{m+3}b^{n+2}+b^{2\left(n+2\right)}\\=a^{2m+6}+2a^{m+3}b^{n+2}+b^{2n+4}$

$\left(2r^{x+1}-4s^{x-1}\right)^2\\=\left(2r^{x+1}\right)^2-2\cdot \:2r^{x+1}\cdot \:4s^{x-1}+\left(4s^{x-1}\right)^2\\=\left(2r^{x+1}\right)^2-16r^{x+1}s^{x-1}+\left(4s^{x-1}\right)^2\\=4r^{2\left(x+1\right)}-16r^{x+1}s^{x-1}+16s^{2\left(x-1\right)}$

$\left(\sqrt{5}a^{2n}-\frac{1}{2}b^{3m}\right)^2\\=\left(\sqrt{5}a^{2n}\right)^2-2\sqrt{5}a^{2n}\frac{1}{2}b^{3m}+\left(\frac{1}{2}b^{3m}\right)^2\\=\frac{20a^{4n}-\sqrt{5}\cdot \:4a^{2n}b^{3m}+b^{6m}}{4}$