Respuesta:
El cuadrado de binomio se calcula de la manera:
[tex](a+b)^2 = a^2 +2ab+b^2[/tex]
Explicación paso a paso:
[tex]a) (9+4m)^2 = 9^2+2\cdot \:9\cdot \:4m+\left(4m\right)^2 = 81+72m+16m^2\\\\b) \left(x^{10}+5y^2\right)^2=\left(x^{10}\right)^2+2x^{10}\cdot \:5y^2+\left(5y^2\right)^2 =x^{20}+10x^{10}y^2+25y^4\\\\c) \left(2x+3z\right)^2=\left(2x\right)^2+2\cdot \:2x\cdot \:3z+\left(3z\right)^2=4x^2+12xz+9z^2\\\\d) \left(4m^5+5n^3\right)^2=\left(4m^5\right)^2+2\cdot \:4m^5\cdot \:5n^3+\left(5n^3\right)^2=16m^{10}+40m^5n^3+25n^6\\[/tex]
[tex]e) \left( \frac{3}{6}w-\frac{1}{2}y \right)^2 = \left( \frac{3}{6}w \right)^2 - 2\cdot \frac{3}{6}w\cdot \frac{1}{2}y+\left(\frac{1}{2}y \right)^2= \frac{9}{36}w^2- \frac{6}{12}wy+ \frac{1}{4}y^2 = \frac{1}{4}w^2- \frac{1}{2}wy+ \frac{1}{4}y^2[/tex]
[tex]f) \left(\frac{5}{7}a^2+\frac{1}{8}n \right)^2 = \left( \frac{5}{7}a^2 \right)^2 + 2\cdot \frac{5}{7}a^2\cdot \frac{1}{8}n + \left( \frac{1}{8}n \right)^2 = \frac{25}{49}a^4 + \frac{5}{28}a^2n+ \frac{1}{64}n^2[/tex]
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Respuesta:
El cuadrado de binomio se calcula de la manera:
[tex](a+b)^2 = a^2 +2ab+b^2[/tex]
Explicación paso a paso:
[tex]a) (9+4m)^2 = 9^2+2\cdot \:9\cdot \:4m+\left(4m\right)^2 = 81+72m+16m^2\\\\b) \left(x^{10}+5y^2\right)^2=\left(x^{10}\right)^2+2x^{10}\cdot \:5y^2+\left(5y^2\right)^2 =x^{20}+10x^{10}y^2+25y^4\\\\c) \left(2x+3z\right)^2=\left(2x\right)^2+2\cdot \:2x\cdot \:3z+\left(3z\right)^2=4x^2+12xz+9z^2\\\\d) \left(4m^5+5n^3\right)^2=\left(4m^5\right)^2+2\cdot \:4m^5\cdot \:5n^3+\left(5n^3\right)^2=16m^{10}+40m^5n^3+25n^6\\[/tex]
[tex]e) \left( \frac{3}{6}w-\frac{1}{2}y \right)^2 = \left( \frac{3}{6}w \right)^2 - 2\cdot \frac{3}{6}w\cdot \frac{1}{2}y+\left(\frac{1}{2}y \right)^2= \frac{9}{36}w^2- \frac{6}{12}wy+ \frac{1}{4}y^2 = \frac{1}{4}w^2- \frac{1}{2}wy+ \frac{1}{4}y^2[/tex]
[tex]f) \left(\frac{5}{7}a^2+\frac{1}{8}n \right)^2 = \left( \frac{5}{7}a^2 \right)^2 + 2\cdot \frac{5}{7}a^2\cdot \frac{1}{8}n + \left( \frac{1}{8}n \right)^2 = \frac{25}{49}a^4 + \frac{5}{28}a^2n+ \frac{1}{64}n^2[/tex]