---
tags: Mundo Mejor, 3ro de Secundaria, 2020
---

<style>
.markdown-body .mark, .markdown-body mark{
background: red;
border-radius: 8px;
color: white;
font-weight: bold;
padding: 5px 10px;
}

h2 {
color: brown;
}
</style>

# Productos notables

## Ejercicio 1


==Solución==

Usando binomio al cuadrado: $\boxed{(a+b)^2 = a^2 + 2ab + b^2}$

Reemplazando:
$$
%%%%%%%%%% Definimos lo necesario %%%%%%%%%%
%definiendo lo necesario
\require{cancel}
\require{color}
\newcommand\ccancel[2][red]{
{\color{#1} \cancel{ \color{black} #2 } }
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align*}
\left ( \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}} \right )^2
&= \left( \sqrt{2+\sqrt{3}} \right)^2 + 2{\left(\sqrt{2+\sqrt{3}}\right) \left(\sqrt{2-\sqrt{3}} \right)} + \left( \sqrt{2-\sqrt{3}} \right)^2 \\[10pt]
&= 2+\sqrt{3} + 2\underbrace{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}_{diferencia\ de\ cuadrados} + 2 - \sqrt{3} \\[10pt]
&= 2+\ccancel{\sqrt{3}} + 2\sqrt{2^2 - \sqrt{3}^2} + 2 - \ccancel{\sqrt{3}} \\[10pt]
&= 4 + 2\sqrt{4-3}\\[10pt]
&= 4 + 2 \\[10pt]
&= 6
\end{align*}
$$

==Rpta: a==

## Ejercicio 2


==Solución==

Aplicando binomio al cuadrado:
$$
\begin{align*}
&= \left[(3x)^2 + 2(3x)(5) + 5^2 \right] - \left[(2x)^2 - 2(2x)(3) + 3^2 \right] - \left( 5x^2 + 42x + 15 \right) \\[10pt]
&= 9x^2 + 30x + 25 - \left[ 4x^2-12x+9 \right] - \left( 5x^2 + 42x + 15 \right) \\[10pt]
&= \ccancel{9x^2} + \ccancel[green]{30x} + 25 - \ccancel{4x^2} + \ccancel[green]{12x} - 9 - \ccancel{5x^2} - \ccancel[green]{42x} - 15 \\[10pt]
&= 25 - 24 \\[10pt]
&= 1
\end{align*}
$$

==Rpta: a==

## Ejercicio 3


==Solución==

Aplicando binomio al cuadrado:
$$
\begin{align*}
&= \dfrac{ \left[ x^2 + 2(x)(2y) + (2y)^2 {\ \Large +\ } (2x)^2 -2(2x)(y) + y^2 -5y^2 \right] }{5} \\[10pt]
&= \dfrac{ x^2 + \ccancel[blue]{4xy} + \ccancel{4y^2} {\ \Large +\ } 4x^2 - \ccancel[blue]{4xy} + \ccancel{y^2} - \ccancel{5y^2} }{5} \\[10pt]
&= \dfrac{\ccancel[green]{5}x^2}{\ccancel[green]{5}} \\[10pt]
&= x^2
\end{align*}
$$

==Rpta: c==

## Ejercicio 4


==Solución==

Elevamos al cuadrado el dato:
$$
\begin{align*}
\left( x + \frac 1x \right)^2 &= (4)^2 \\[10pt]
x^2 + 2 \cdot \ccancel{x} \cdot \frac{1}{\ccancel{x}} + \left( \frac 1x \right)^2 &= 16\\[10pt]
x^2 + 2 + \frac {1}{x^2} &= 16 \\[10pt]
x^2 + \frac {1}{x^2} &= 14
\end{align*}
$$

==Rpta: c==

## Ejercicio 5


==Solución==

Elevamos al cuadrado el dato:
$$
\begin{align*}
\left( x + \frac 1x \right)^2 &= (5)^2 \\[10pt]
x^2 + 2 \cdot \ccancel{x} \cdot \frac{1}{\ccancel{x}} + \left( \frac 1x \right)^2 &= 25\\[10pt]
x^2 + 2 + \frac {1}{x^2} &= 25 \\[10pt]
x^2 + \frac {1}{x^2} &= 23
\end{align*}\\
$$

Elevamos nuevamente al cuadrado:
$$
\begin{align*}
\left(x^2 + \frac {1}{x^2}\right)^2 &= (23)^2 \\[10pt]
x^4 + 2 \cdot \ccancel{x^2} \cdot \frac{1}{\ccancel{x^2}} + \left( \frac{1}{x^2} \right)^2 &= 529\\[10pt]
x^4 + 2 + \frac {1}{x^4} &= 529 \\[10pt]
x^4 + \frac {1}{x^4} &= 527
\end{align*}
$$

==Rpta: d==

## Ejercicio 6


==Solución==

Usamos Legendre: $\boxed{(a+b)^2-(a-b)^2=4ab}$

Entonces:
$$
\begin{align*}
2x^2+2y^2 &= \underbrace{(x+y)^2-(x-y)^2} \\[10pt]
2x^2+2y^2 &= 4xy
\end{align*}
$$

Simplificando toda la expresión:
$$
\ccancel{2}x^2+\ccancel{2}y^2 = \ccancel{4}xy
$$

Ordenando términos:
$$
\begin{align*}
x^2+y^2 &= 2xy \\[10pt]
\underbrace{x^2 -2xy +y^2}_{trinomio\ cuadrado\ perfecto} &= 0 \\[10pt]
(x-y)^2 &= 0 \\[10pt]
x - y &= 0 \implies \boxed{x = y}
\end{align*}
$$

Reemplazando en lo pedido:
$$
\begin{align*}
\left( \frac xy \right)^{10} &= \left( \frac xx \right)^{10} \\[10pt]
&= (1)^{10} = 1
\end{align*}
$$

==Rpta: b==

## Ejercicio 7


==Solución==

Dando forma a la expresión:
$$
\large
=\sqrt[32]{{\color{blue}(2^4-1)}(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)+1}
$$

Aplicamos diferencia de cuadrados: $\boxed{(a+b)(a-b)=a^2-b^2}$
$$
\large
\begin{align*}
&=\sqrt[32]{\underbrace{(2^4-1)(2^4+1)}(2^8+1)(2^{16}+1)(2^{32}+1)+1} \\[10pt]
&=\sqrt[32]{\underbrace{(2^8-1)(2^8+1)}(2^{16}+1)(2^{32}+1)+1}\\[10pt]
&=\sqrt[32]{\underbrace{(2^{16}-1)(2^{16}+1)}(2^{32}+1)+1}\\[10pt]
&=\sqrt[32]{\underbrace{(2^{32}-1)(2^{32}+1)}+1}\\[10pt]
&=\sqrt[32]{2^{64}-\ccancel{1}+\ccancel{1}}\\[10pt]
&=\sqrt[\ccancel{32}]{2^{\ccancel{64}}}\\[10pt]
&=2^2=4
\end{align*}
$$

==Rpta: d==

## Ejercicio 8


==Solución==

Dando forma a la expresión:
$$
\begin{align*}
Q &= ({\color{blue} 5^2 - 3^2})(5^2+3^2)(5^4+3^4)(5^8+3^8)(5^{16}+3^{16})+3^{32}
\end{align*}\\
$$

Aplicamos diferencia de cuadrados: $\boxed{(a+b)(a-b)=a^2-b^2}$

$$
\begin{align*}
Q &= \underbrace{(5^2 - 3^2)(5^2+3^2)}(5^4+3^4)(5^8+3^8)(5^{16}+3^{16})+3^{32} \\[10pt]
&= \underbrace{(5^4-3^4)(5^4+3^4)}(5^8+3^8)(5^{16}+3^{16})+3^{32} \\[10pt]
&= \underbrace{(5^8-3^8)(5^8+3^8)}(5^{16}+3^{16})+3^{32} \\[10pt]
&= \underbrace{(5^{16}-3^{16})(5^{16}+3^{16})}+3^{32} \\[10pt]
&= 5^{32}-\ccancel{3^{32}} + \ccancel{3^{32}} \\[10pt]
&= 5^{32} \\[10pt]
\end{align*}\\
$$

==Rpta: c==

## Ejercicio 9


==Solución==

Elevamos al cuadrado el dato:
$$
\begin{align*}
\left( x + \frac 1x \right)^2 &= (3)^2 \\[10pt]
x^2 + 2 \cdot \ccancel{x} \cdot \frac{1}{\ccancel{x}} + \left( \frac 1x \right)^2 &= 9\\[10pt]
x^2 + 2 + \frac {1}{x^2} &= 9 \\[10pt]
x^2 + \frac {1}{x^2} &= 7
\end{align*}
$$

Elevamos al cubo:
$$
\begin{align*}
\left( x^2 + \frac{1}{x^2} \right)^3 &= (7)^3 \\[10pt]
(x^2)^3+\left(\frac{1}{x^2}\right)^3+3\left(\ccancel{x^2}\right)\left(\frac{1}{\ccancel{x^2}}\right)\left(\underbrace{x^2+\frac{1}{x^2}}_7\right) &= 343 \\[10pt]
x^6+\frac{1}{x^6}+3(1)(7) &= 343 \\[10pt]
x^6+\frac{1}{x^6} &= 343-21 \\[10pt]
x^6+\frac{1}{x^6} &= 322 \\[10pt]
\end{align*}
$$

==Rpta: d==

## Ejercicio 10


==Solución==

Aplicamos diferencia de cuadrados: $\boxed{(a+b)(a-b)=a^2-b^2}$

$$
\begin{align*}
E &= (987654322+987654320)(987654322-987654320) \\[10pt]
E &= (1975308642)(2) \\[10pt]
E &= 3950617284‬
\end{align*}
$$