Question

Se A= 3^x + 3^-x sobre 2 e B= 3^x - 3^-x sobre 2, qual o valor de A² - B²?

Me ajudem por favor!

Answer 1

A= 3x +3-x
B=3x-3- x
…A² - B² = (A + B)(A - B)
A² - B² = (3x + 3 –…A² - B² = (3x – x + 3x – x + 3 – 3)[3x + 3 – x – 3x + 3 + x]
…A² - B² = 4x[3x – 3x + x – x + 3 + 3]
…A² - B² = 4x∙6
⇨ A² - B² = 24x 
x + 3x – 3 – x)[3x + 3 – x – (3x – 3 – x)]

Answer 2

Olá,

use as propriedades da exponenciação:

$\text{A}^2-\text{B}^2=\left( \dfrac{3^x+3^{-x}}{2}\right)^2-\left( \dfrac{3^x-3^{-x}}{2} \right)^2\\\\\\ \text{A}^2-\text{B}^2=\left[ \dfrac{3^{2\cdot x}+3^{2\cdot(-x)}}{2^2}\right]-\left[ \dfrac{3^{2\cdot x}-3^{2\cdot(-x)}}{2^2} \right]\\\\\\ \text{A}^2-\text{B}^2=\left( \dfrac{3^{2x}+3^{-2x}}{4}\right)-\left( \dfrac{3^{2x}-3^{-2x}}{4} \right)$

$\text{A}^2-\text{B}^2= \dfrac{3^{2x}}{4}+ \dfrac{3^{-2x}}{4}- \dfrac{3^{2x}}{4}+ \dfrac{3^{-2x}}{4}\\\\\\ \text{A}^2-\text{B}^2= \dfrac{3^{-2x}}{4}+\dfrac{3^{-2x}}{4}\\\\\\ \Large\boxed{ \text{A}^2-\text{B}^2= 2\cdot\left(\dfrac{3^{-2x}}{4}\right)}$

Tenha ótimos estudos ;D