Nos piden resolver los cuadrados de sumas de dos términos:
Fórmula del binomio al cuadrado:
[tex]\bf{\left(a+b\right)^2=a^2+2ab+b^2}\\\\\left(a-b\right)^2=a^2-2ab+b^2[/tex]
Entonces:
a) [tex]\left(2m+n\right)^2[/tex]
[tex]=\left(2m\right)^2+2\cdot \:(2m)(n)+(n)^2\\\\\boxed{=4m^2+4mn+n^2}[/tex]
b) [tex]\left(5xy+3x^2\right)^2[/tex]
[tex]\left(5xy+3x^2\right)^2=\left(5xy\right)^2+2\cdot \:(5xy)\cdot \:(3x^2)+\left(3x^2\right)^2\\\\=\left(5xy\right)^2+2\cdot \:(5xy)\cdot \:(3x^2)+\left(3x^2\right)^2\\\\\boxed{=25x^2y^2+30x^3y+9x^4}[/tex]
c) [tex]\left(8+\frac{2}{7}m\right)^2[/tex]
[tex]\left(8+\frac{2}{7}m\right)^2=8^2+2\left(8\right)\left(\frac{2}{7}m\right)+\left(\frac{2}{7}m\right)^2\\\\=64+16\cdot \frac{2}{7}m+\frac{4m^2}{49}\\\\=\frac{4m^2}{49}+16\cdot \frac{2}{7}m+64\\\\=\left(\frac{2}{7}m\right)^2+16\cdot \frac{2}{7}m+8^2\\\\=\left(\frac{2}{7}m+8\right)^2\\\\=\left(\frac{56+2m}{7}\right)^2\\\\=\frac{\left(56+2m\right)^2}{7^2}\\\\\boxed{=\frac{\left(56+2m\right)^2}{49}}[/tex]
d) [tex]\left(3m^2n^3\right)^2[/tex]
[tex]\left(3m^2n^3\right)^2=3^2\left(m^2\right)^2\left(n^3\right)^2\\\\=3^2\left(m^2\right)^2\left(n^3\right)^2\\\\=9\left(m^2\right)^2\left(n^3\right)^2\\\\=9m^4\left(n^3\right)^2\\\\\boxed{=9m^4n^6}[/tex]
e) [tex]\left(\frac{7}{8}m^5+1.5\right)^2[/tex]
[tex]=\left(\frac{7m^5}{8}+1.5\right)^2\\\\\frac{7}{8}=0.875\\\\=\left(0.875m^5+1.5\right)^2\\\\\left(0.875m^5+1.5\right)^2=\left(0.875m^5\right)^2+2\cdot \:0.875m^5\cdot \:1.5+1.5^2\\\\=\left(0.875m^5\right)^2+2\cdot \:0.875m^5\cdot \:1.5+1.5^2\\\\\boxed{=0.765625m^{10}+2.625m^5+2.25}[/tex]
f) [tex]\left(\frac{1}{2}P^{x+2}+\frac{1}{5}q^{zy-1}\right)^2[/tex]
[tex](\frac{p^{z+2}}{2} +\frac{1q^{zy-1}}{5})^2[/tex]
[tex]=\left(\frac{5P^{x+2}+2q^{zy-1}}{10}\right)^2\\\\=\frac{\left(5P^{x+2}+2q^{zy-1}\right)^2}{10^2}\\\\\boxed{=\frac{\left(5P^{x+2}+2q^{zy-1}\right)^2}{100}}[/tex]
MUCHA SUERTE...