Matemáticas 1 · Tema 5
Expresiones algebraicas
Ejemplos
$$\begin{aligned}
4[(+3x-2)(5-3x)+2x]-4 &= \\
4[15x-9x^2-10+6x+2x]-4 &= \\
60x-36x^2-40+24x+8x-4 &= \\
\mathbf{-36x^2+92x-44} &
\end{aligned}$$
$$\begin{aligned}
2(-3x+1)^2+2(-4+3x)-2 &= \\
2(-3x+1)(-3x+1)-8+6x-2 &= \\
2(9x^2-3x-3x+1)-8+6x-2 &= \\
2(9x^2-6x+1)-8+6x-2 &= \\
18x^2-12x+2-8+6x-2 &= \\
\mathbf{18x^2-6x-8} &
\end{aligned}$$
Ejercicios
Bloque 1
$4[(-4x+5)(2-5x)+6x]-3=$
$4(-3x+2)^2+3(-1+5x)-6=$
$2[(-3x+2)(4-5x)+3x]-5=$
$5(2x-1)^2+4(-3+6x)-3=$
Bloque 2
$3[(-2x+3)(5-4x)+2x]-5=$
$5(-4x+3)^2-2(3-2x)-4=$
$5[(2x-3)(1-3x)+4x]-2=$
$2(-2x+4)^2+3(-5x+3)-1=$
Ejercicios resueltos
▼
$$\begin{aligned}
4[(-4x+5)(2-5x)+6x]-3 &= \\
4[-8x+20x^2+10-25x+6x]-3 &= \\
-32x+80x^2+40-100x+24x-3 &= \\
\mathbf{80x^2-108x+37} &
\end{aligned}$$
$$\begin{aligned}
4(-3x+2)^2+3(-1+5x)-6 &= \\
4(-3x+2)(-3x+2)-3+15x-6 &= \\
4(9x^2-6x-6x+4)-3+15x-6 &= \\
4(9x^2-12x+4)-3+15x-6 &= \\
36x^2-48x+16-3+15x-6 &= \\
\mathbf{36x^2-33x+7} &
\end{aligned}$$
$$\begin{aligned}
2[(-3x+2)(4-5x)+3x]-5 &= \\
2[-12x+15x^2+8-10x+3x]-5 &= \\
-24x+30x^2+16-20x+6x-5 &= \\
\mathbf{30x^2-38x+11} &
\end{aligned}$$
$$\begin{aligned}
5(2x-1)^2+4(-3+6x)-3 &= \\
5(2x-1)(2x-1)-12+24x-3 &= \\
5(4x^2-2x-2x+1)-12+24x-3 &= \\
5(4x^2-4x+1)-12+24x-3 &= \\
20x^2-20x+5-12+24x-3 &= \\
\mathbf{20x^2+4x-10} &
\end{aligned}$$
$$\begin{aligned}
3[(-2x+3)(5-4x)+2x]-5 &= \\
3[-10x+8x^2+15-12x+2x]-5 &= \\
-30x+24x^2+45-36x+6x-5 &= \\
\mathbf{24x^2-60x+40} &
\end{aligned}$$
$$\begin{aligned}
5(-4x+3)^2-2(3-2x)-4 &= \\
5(-4x+3)(-4x+3)-6+4x-4 &= \\
5(16x^2-12x-12x+9)-6+4x-4 &= \\
5(16x^2-24x+9)-6+4x-4 &= \\
80x^2-120x+45-6+4x-4 &= \\
\mathbf{80x^2-116x+35} &
\end{aligned}$$
$$\begin{aligned}
5[(2x-3)(1-3x)+4x]-2 &= \\
5[2x-6x^2-3+9x+4x]-2 &= \\
10x-30x^2-15+45x+20x-2 &= \\
\mathbf{-30x^2+75x-17} &
\end{aligned}$$
$$\begin{aligned}
2(-2x+4)^2+3(-5x+3)-1 &= \\
2(-2x+4)(-2x+4)-15x+9-1 &= \\
2(4x^2-8x-8x+16)-15x+9-1 &= \\
2(4x^2-16x+16)-15x+9-1 &= \\
8x^2-32x+32-15x+9-1 &= \\
\mathbf{8x^2-47x+40} &
\end{aligned}$$
Ejercicios extra
▼Bloque A
$3[(2x+3)(4-x)+5x]-2=$
$2(x-4)^2+3(2x+5)-7=$
$4[(-3x+1)(5-2x)+3x]+1=$
$4(3x-1)^2+2(2-5x)-5=$
Bloque B
$2[(x+5)(3-2x)+4x]-3=$
$3(2x+5)^2-4(3-x)+2=$
$5[(-x+3)(2-4x)+2x]-1=$
$4(-x+2)^2+5(-3+4x)-6=$