Question

alguien me me ayudes con un par?​

Answer 1

Verified answer

RESOLVER:

a) [tex]\left(5x^3zb+7d^5e^7\right)^2[/tex]

[tex]=\left(5bx^3z+7e^7d^5\right)^2[/tex]

[tex]a=5x^3zb,\:\:b=7d^5e^7[/tex]

[tex]=\left(5x^3zb\right)^2+2\cdot \:5x^3zb\cdot \:7d^5e^7+\left(7d^5e^7\right)^2[/tex]

[tex]=\left(5x^3zb\right)^2+2\cdot \:5x^3zb\cdot \:7d^5e^7+\left(7d^5e^7\right)^2: 25b^2x^6z^2+70e^7bd^5x^3z+49e^{14}d^{10}[/tex]

[tex]=25b^2x^6z^2+70e^7bd^5x^3z+49e^{14}d^{10}[/tex]

b) [tex]\left(3b+5d\right)^2[/tex]

[tex]\left(3b+5d\right)^2=\left(3b\right)^2+2\cdot \:3b\cdot \:5d+\left(5d\right)^2[/tex]

[tex]=\left(3b\right)^2+2\cdot \:3b\cdot \:5d+\left(5d\right)^2[/tex]

[tex]\left(3b\right)^2+2\cdot \:3b\cdot \:5d+\left(5d\right)^2:9b^2+30bd+25d^2[/tex]

[tex]=9b^2+30bd+25d^2[/tex]

c) [tex]\left(\frac{4}{5}ax^2b-\frac{1}{6}x\right)^2[/tex]

[tex]\frac{4}{5}ax^2b=\frac{4abx^2}{5}[/tex]

[tex]\frac{1}{6}x=\frac{x}{6}[/tex]

[tex]=\left(\frac{4abx^2}{5}-\frac{x}{6}\right)^2[/tex]

d) [tex]\left(\frac{1}{7}a^2c-\frac{4}{3}x\right)^2[/tex]

[tex]\frac{1}{7}a^2c:\frac{a^2c}{7}[/tex]

[tex]\frac{4}{3}x:\frac{4x}{3}[/tex]

[tex]=\left(\frac{a^2c}{7}-\frac{4x}{3}\right)^2[/tex]

EJERCICIOS DE BINOMIOS CONJUGADOS:

a) [tex]\left(7+6\right)\left(7-6\right)[/tex]

[tex]\left(7+6\right):13[/tex]

[tex]\left(7-6\right):1[/tex]

[tex]=13\cdot \:1[/tex]

[tex]=13[/tex]

b) [tex]\left(\frac{2}{3}p^5+\frac{4}{5}r^4\right)\left(\frac{2}{3}p^5-\frac{4}{5}r^4\right)[/tex]

[tex]\frac{2}{3}p^5:\frac{2p^5}{3}[/tex]

[tex]\frac{4}{5}r^4: \frac{4r^4}{5}[/tex]

[tex]\frac{2}{3}p^5:\frac{2p^5}{3}[/tex]

[tex]\frac{4}{5}r^4:\frac{4r^4}{5}[/tex]

[tex]=\left(\frac{4r^4}{5}+\frac{2p^5}{3}\right)\left(-\frac{4r^4}{5}+\frac{2p^5}{3}\right)[/tex]

[tex]\frac{4p^{10} }{10} -\frac{16r^8}{25}[/tex]

c) [tex]\left(4a^3b^2+5x^4y^6\right)\left(4a^3b^2-5x^4y^6\right)[/tex]

[tex]\left(4a^3b^2+5x^4y^6\right)\left(4a^3b^2-5x^4y^6\right)=\left(4a^3b^2\right)^2-\left(5x^4y^6\right)^2[/tex]

[tex]\left(4a^3b^2\right)^2-\left(5x^4y^6\right)^2:16a^6b^4-25x^8y^{12}[/tex]

[tex]=16a^6b^4-25x^8y^{12}[/tex]

d) [tex]\left(2x+3a\right)\left(2x-3a\right)[/tex]

[tex]\left(2x+3a\right)\left(2x-3a\right)=\left(2x\right)^2-\left(3a\right)^2[/tex]

[tex]=\left(2x\right)^2-\left(3a\right)^2[/tex]

[tex]\left(2x\right)^2-\left(3a\right)^2:4x^2-9a^2[/tex]

[tex]=4x^2-9a^2[/tex]

e) [tex]\left(5b^3c^5+6xy\right)\left(5b^3c^5-6xy\right)[/tex]

[tex]=\left(5b^3c^5\right)^2-\left(6xy\right)^2[/tex]

[tex]=25b^6c^{10}-36x^2y^2[/tex]

f) [tex]\left(6ab^2+8bd^2\right)\left(6ab^2-8bd^2\right)[/tex]

[tex]=\left(6ab^2\right)^2-\left(8bd^2\right)^2[/tex]

[tex]=36a^2b^4-64b^2d^4[/tex]

g) [tex]\left(x^3+3y\right)\left(x^3-3y\right)[/tex]

[tex]=\left(x^3\right)^2-\left(3y\right)^2[/tex]

[tex]=x^6-9y^2[/tex]

h) [tex]\left(x^2+4x^3\right)\left(x^2-4x^3\right)[/tex]

[tex]=\left(x^2\right)^2-\left(4x^3\right)^2[/tex]

[tex]=x^4-16x^6[/tex]

[tex]Buena\:suerte\:con\:tus\:tareas\::ProfeAndresFelipe :)[/tex]