Name: ▶️ Ejercicio 93 álgebra - Solucionario Baldor ✅✅
Item: ▶️ Ejercicio 93 álgebra - Solucionario Baldor ✅✅
Rating: 5
Author: Gestor Baldor

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DESCARGA

CAPITULO X

Descomposición Factorial

Ejercicio 93

Factorar o descomponer en dos factores:

$x^{2} - y^{2} = (x + y) (x - y)$
$a^{2} - 1 = (a + 1) (a - 1)$
$a^{2} - 4 = (a + 2) (a - 2)$
$9 - b^{2} = (3 + b) (3 - b)$
$1 - 4 m^{2} = (1 - 2 m) (1 + 2 m)$
$16 - n^{2} = (4 - n) (4 + n)$
$a^{2} - 25 = (a + 5) (a - 5)$
$1 - y^{2} = (1 + y) (1 - y)$
$4 a^{2} - 9 = (2 a - 3) (2 a + 3)$
$25 - 36 x^{4} = (5 + 6 x^{2}) (5 - 6 x^{2})$
$1 - 49 a^{2} b^{2} = (1 + 7 a b) (1 - 7 a b)$
$4 x^{2} - 81 y^{4} = (2 x + 9 y^{2}) (2 x - 9 y^{2})$
$a^{2} b^{8} - c^{2} = (a b^{4} + c) (a b^{4} - c)$
$100 - x^{2} y^{6} = (10 - x y^{3}) (10 + x y^{3})$
$a^{10} - 49 b^{12} = (a^{5} - 7 b^{6}) (a^{5} + 7 b^{6})$
$25 x^{2} y^{4} - 121 = (5 x y^{2} + 11) (5 x y^{2} - 11)$
$100 m^{2} n^{4} - 169 y^{6} = (10 m n^{2} - 13 y^{3}) (10 m n^{2} + 13 y^{3})$
$a^{2} m^{4} n^{6} - 144 = (a m^{2} n^{3} - 12) (a m^{2} n^{3} + 12)$
$196 x^{2} y^{4} - 225 z^{12} = (14 x y^{2} - 15 z^{6}) (14 x y^{2} + 15 z^{6})$
$256 a^{12} - 289 b^{4} m^{10} = (16 a^{6} - 17 b^{2} m^{5}) (16 a^{6} + 17 b^{2} m^{5})$
$1 - 9 a^{2} b^{4} c^{6} d^{8} = (1 + 3 a b^{2} c^{3} d^{4}) (1 - 3 a b^{2} c^{3} d^{4})$
$361 x^{14} - 1 = (19 x^{7} + 1) (19 x^{7} - 1)$
$\frac{1}{4} - 9 a^{2} = (\frac{1}{2} - 3 a) (\frac{1}{2} + 3 a)$
$1 - \frac{a^{2}}{25} = (1 + \frac{a}{5}) (1 - \frac{a}{5})$
$\frac{1}{16} - \frac{4 x^{2}}{49} = (\frac{1}{4} - \frac{2 x}{7}) (\frac{1}{4} + \frac{2 x}{7})$
$\frac{a^{2}}{36} - \frac{x^{6}}{25} = (\frac{a}{6} - \frac{x^{3}}{5}) (\frac{a}{6} + \frac{x^{3}}{5})$
$\frac{x^{2}}{100} - \frac{y^{2} z^{4}}{81} = (\frac{x}{10} - \frac{y z^{2}}{9}) (\frac{x}{10} + \frac{y z^{2}}{9})$
$\frac{x^{6}}{49} - \frac{4 a^{10}}{121} = (\frac{x^{3}}{7} - \frac{2 a^{5}}{11}) (\frac{x^{3}}{7} + \frac{2 a^{5}}{11})$
$100 m^{2} n^{4} - \frac{1}{16} x^{8} = (10 m n^{2} - \frac{1}{4} x^{4}) (10 m n^{2} + \frac{1}{4} x^{4})$
$a^{2 n} - b^{2 n} = (a^{n} + b^{n}) (a^{n} - b^{n})$
$4 x^{2 n} - \frac{1}{9} = (2 x^{n} - \frac{1}{3}) (2 x^{n} + \frac{1}{3})$
$a^{4 n} - 225 b^{4} = (a^{2 n} + 15 b^{2}) (a^{2 n} - 15 b^{2})$
$16 x^{6 m} - \frac{y^{2 n}}{49} = (4 x^{3 m} - \frac{y^{n}}{7}) (4 x^{3 m} + \frac{y^{n}}{7})$
$49 a^{10 n} - \frac{b^{12 x}}{81} = (7 a^{5 n} - \frac{b^{6 x}}{9}) (7 a^{5 n} + \frac{b^{6 x}}{9})$
$a^{2 n} b^{4 n} - \frac{1}{25} = (a^{n} b^{2 n} - \frac{1}{5}) (a^{n} b^{2 n} + \frac{1}{5})$
$\frac{1}{100} - x^{2 n} = (\frac{1}{10} - x^{n}) (\frac{1}{10} + x^{n})$