En los problemas del 1 al 14, resolver las ecuaciones dadas.
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\[ 5(x – 3) = 3(x + 7) + x \]
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\[ y – (6 – 2y) = 8(y – 2) \]
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\[ \frac{1}{2} (2x-1) = 3\left(x+\frac{1}{4} \right) \]
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\[ \frac{x}{4} – \frac{x}{3} = \frac{7}{6} – \frac{4x}{3} \]
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\[ \frac{2x-1}{5} = \frac{2+x}{3} \]
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\[ \frac{7z+1}{6} + \frac{3}{2} = \frac{3z}{4} \]
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\[ \frac{x-1}{3} – \frac{2-3x}{14} = \frac{4x-3}{7} \]
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\[ \frac{x-3}{6} – \frac{2x-1}{5} =-1 \]
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\[ \frac{x+1}{5} + \frac{x+2}{6} = \frac{x-1}{4} + \frac{x+7}{10} \]
\[ \begin{aligned} &\frac{x+1}{5} + \frac{x+2}{6} \\[1em] &\hspace{3em}= \frac{x-1}{4} + \frac{x+7}{10} \end{aligned} \]
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\[ \frac{5x-2}{3} – \frac{1}{2} (3x-1) = \frac{9x+7}{6} – \frac{2}{9} (5x-1) \]
\[ \begin{aligned} &\frac{5x-2}{3} – \frac{1}{2} (3x-1) \\[1em] &\hspace{3em}= \frac{9x+7}{6} – \frac{2}{9} (5x-1) \end{aligned} \]
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\[ (x-3)^2 = (x-1)^2 \]
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\[ (x-5)(x+1) = (x+2)(x-3) + 13 \]
\[ \begin{aligned} &(x-5)(x+1) \\[1em] &\hspace{4em} = (x+2)(x-3) + 13 \end{aligned} \]
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\[ (2x-5) (x-1) + x^2 = (3x-1) (x+2) +1 \]
\[ \begin{aligned} &(2x-5) (x-1) + x^2 \\[1em] &\hspace{3em} = (3x-1) (x+2) +1 \end{aligned} \]
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\[ 8x(x+2) (x-1) = (2x+1)^3 – (2x+3)^2 \]
\[ \begin{aligned} &8x(x+2) (x-1) \\[1em] &\hspace{3em} = (2x+1)^3 – (2x+3)^2 \end{aligned} \]
En los problemas del 15 al 22, resolver las ecuaciones (despejar \(\boldsymbol{x}\)).
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\[ 5(5x-a) = a^2 (x-1) \]
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\[ a(x+b) + x(b-a) = 2b (2a-x) \]
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\[ x^2 + b^2 + b(b-1) = (x+b)^2 \]
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\[ (x+a)^3 -2x^3 = 12a^3 – (x-a) \]
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\[ \frac{x-a}{b} + \frac{x-b}{a} = 2 \]
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\[ \frac{x-3m}{m^2} + \frac{x-2m}{mn} = -\frac{1}{m} \]
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\[ \frac{a-x}{a} – \frac{b-x}{b} = \frac{2(a-b)}{ab} \]
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\[ \frac{x-a}{a+b} + \frac{a+b}{a-b} = \frac{x+b}{a+b} + \frac{x-b}{a-b} \]
En los problemas del 23 al 26 despejar la variable indicada en términos de las otras.
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\[ A = \pi (r^2+rs) ,\; s \]
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\[ S = a\frac{1-r^n}{1-r} ,\; a \]
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\[ S = \frac{f}{H-h} ,\; h \]
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\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{a} ,\; x \]
En los problemas del 27 al 40, resolver las ecuaciones dadas factorizando.
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\[ x^2-4x-12=0 \]
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\[ x^2-6x+9=0 \]
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\[ x^2+24=-11x \]
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\[ 2x^2-3x+1=0 \]
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\[ 9x^2-17x-2=0 \]
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\[ (2x-1)^2 – (x+5)^2 =-19 \]
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\[ (x-5)^2 – (x-4)^2 = (2x+3)^2 +12 \]
\[ \begin{aligned} &(x-5)^2 – (x-4)^2 \\[1em] &\hspace{5em}= (2x+3)^2 +12 \end{aligned} \]
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\[ (x-2)^3 – (x+1)^3 = -x(3x+4) -24 \]
\[ \begin{aligned} &(x-2)^3 – (x+1)^3 \\[1em] &\hspace{4em}= -x(3x+4) -24 \end{aligned} \]
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\[ 6x^2 -\frac{5x}{2} = -\frac{1}{4} \]
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\[ \frac{2(x+5)}{5} + \frac{x-4}{4} = \frac{x^2-53}{5} \]
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\[ x^4-17x^2+16=0 \]
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\[ 6y^4 = \frac{y^2}{2} + \frac{1}{4} \]
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\[ x^{2/3}+x^{1/3}-6=0 \]
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\[ 2x^{2/3}+3x^{1/3}-2=0 \]
En los problemas del 41 al 46, resolver las ecuaciones dadas, mediante la fórmula cuadrática.
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\[ 9(x-1)^2=5 \]
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\[ \sqrt{3}x-3=4x^2 \]
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\[ 2x(2x-3)=-1 \]
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\[ (x+15)^2=6x(x+5) \]
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\[ x^2-2x-(a^2+2a)=0 \]
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\[ \frac{x^2}{2a} – \frac{a+2}{2a} x+1=0 \]
En los problemas del 47 al 60, resolver las ecuaciones fraccionarias.
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\[ \frac{x-6}{x} = \frac{x+6}{x-6} + \frac{6}{x} \]
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\[ \frac{x}{x+2} – \frac{x}{x-2} = \frac{x-15}{x^2-4} \]
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\[ \frac{1}{3x-3} + \frac{1}{4x+4} = \frac{1}{12x-12} \]
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\[ \frac{4x+1}{4x-1} = \frac{4x-1}{4x+1} + \frac{6}{16x^2-1} \]
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\[ \frac{1}{x} + \frac{1}{4-x} = 1 \]
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\[ \frac{x}{1+x} + \frac{1}{1-x} = 0 \]
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\[ \frac{3y-2}{3y+2} = \frac{2y+3}{4y-1} \]
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\[ \frac{ x+5 }{ (x-1)(x+2) } = \frac{2x}{x+2} \]
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\[ \frac{1}{x-1} – \frac{1}{x-2} = \frac{1}{x-3} \]
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\[ \frac{3x}{x-2} – \frac{1}{x^2-4} = 2 \]
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\[ \frac{1}{x^2} + \frac{2}{x} -15 = 0 \]
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\[ \frac{12}{x-1} + \frac{12}{x} = 10 \]
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\[ \frac{2x}{x-1} = \frac{8}{x-1} – \frac{5}{x} \]
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\[ \frac{1}{x^2-4} + \frac{2x+3}{x+2} + \frac{x+3}{x-2} = 0 \]
En los problemas del 61 al 76, resolver las ecuaciones radicales dadas. Eliminar las soluciones extrañas.
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\[ 5-\sqrt{2x+3}= \]
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\[ \sqrt{ \frac{x}{18} + 1 } = \frac{2}{3} \]
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\[ (5x-1)^{1/2}=7 \]
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\[ (y+9)^{3/2} = 4^3 \]
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\[ \sqrt{x^2-5} = 5-x \]
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\[ \sqrt{z+7} – \sqrt{z} = 1 \]
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\[ \sqrt{ 9x^2 – 10x } = 3x-2 \]
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\[ \sqrt{ \frac{1}{x} } – \sqrt{ \frac{8}{4x+1} } = 0 \]
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\[ \sqrt{4x+1} + 1 = 2x \]
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\[ \sqrt{x^2+5} = 2x-1 \]
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\[ \sqrt{x+5} = 2\sqrt{x} – 1 \]
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\[ \sqrt{x} + \sqrt{x-3} = \sqrt{x+5} \]
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\[ \sqrt{ x + \sqrt{x+8} } = 2\sqrt{x} \]
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\[ \sqrt{ x + \sqrt{x+8} } = 2\sqrt{x} \]
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\[ \sqrt{ x + \sqrt{x+8} } = 2\sqrt{x} \]
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\[ \frac{x}{2} = \frac{ \sqrt{x+2} – \sqrt{x-2} }{ \sqrt{x+2} + \sqrt{x-2} } \]
En los problemas 77 y 78, resolver efectuando un cambio de variable.
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\[ \left( \frac{3x}{x+1} \right)^2 – \frac{6x}{x+1} = 8 \]
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\[ \sqrt[3]{ \frac{5x+4}{x-1} } + \sqrt[3]{ \frac{x-1}{5x+4} } = \frac{5}{2} \]
En los problemas 79 y 80, usando el teorema del residuo, hallar el residuo cuando se divide:
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\[ 3x^4 – 5x^3 – 4x^2 + 3x – 2, \; \text{ entre } (x-2) \]
\[ \begin{aligned} &3x^4 – 5x^3 – 4x^2 + 3x – 2, \\[1em] &\hspace{4em}\text{entre } (x-2) \end{aligned} \]
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\[ x^3 – 6x^2 +11x – 6, \; \text{ entre } (x+2) \]
\[ \begin{aligned} &x^3 – 6x^2 +11x – 6, \\[1em] &\hspace{4em} \text{entre } (x+2) \end{aligned} \]
En los problemas del 81 al 88, hallar las raíces de la ecuación dada y factorice el polinomio correspondiente.
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\[ x^3 + 2x^2 – x – 2 = 0 \]
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\[ x^3 – 3x^2 + 2 = 0 \]
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\[ 4x^3 – 7x^2 + 3 = 0 \]
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\[ 2x^3 – 2x^2 – 11x + 2 = 0 \]
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\[ x^4 – x^3 – 5x^2 + 3x + 6 = 0 \]
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\[ 3x^4 + 5x^3 – 5x^2 – 5x + 2 = 0 \]
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\[ x^5 -3x^4 -5x^3 +15x^2+4x -12 = 0 \]
\[ \begin{aligned} x^5 -3x^4 -5x^3 &+15x^2 \\[1em] &+ 4x -12 = 0 \end{aligned} \]
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\[ x^5 + 4x^4 – 4x^3 -34x^2 -45x-18 = 0 \]
\[ \begin{aligned} x^5 + 4x^4 – 4x^3 &-34x^2 \\[1em] &-45x-18 = 0 \end{aligned} \]
En los problemas del 89 al 91, usar el teorema del factor para probar que:
- \(x – a\) es un factor de \(x^n – a^n\), para todo entero positivo \(n\).
- \(x + a\) es un factor de \(x^n – a^n\), para todo entero positivo par \(n\).
- \(x + a\) es un factor de \(x^n + a^n\) , para todo entero positivo impar \(n\).
