Respuesta: Operamos usando la fórmula del binomio al cuadrado:
[tex](a \pm b)^{2} = a^{2} \pm 2ab + b^{2}[/tex]
En la que:
Basándonos en esto:
[tex](3x + 1)^{2} - (3x - 1)^{2} - 5(2x - 1) = (x + 1)^{2}\\\\(9x^{2} + 6x + 1) - (9x^{2} - 6x + 1) - 10x + 5 = x^{2} + 2x + 1\\\\12x - 10x + 5 = x^{2} + 2x + 1\\\\x^{2} - 4 = 0\\\\x^{2} = 4\\\\x_1 = 2\\x_2 = -2[/tex]
Espero que te ayude ^_^
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Verified answer
Respuesta: Operamos usando la fórmula del binomio al cuadrado:
[tex](a \pm b)^{2} = a^{2} \pm 2ab + b^{2}[/tex]
En la que:
Basándonos en esto:
[tex](3x + 1)^{2} - (3x - 1)^{2} - 5(2x - 1) = (x + 1)^{2}\\\\(9x^{2} + 6x + 1) - (9x^{2} - 6x + 1) - 10x + 5 = x^{2} + 2x + 1\\\\12x - 10x + 5 = x^{2} + 2x + 1\\\\x^{2} - 4 = 0\\\\x^{2} = 4\\\\x_1 = 2\\x_2 = -2[/tex]
Espero que te ayude ^_^