4 Soluciones a las actividades de cada epígrafe PÁGINA 85 Pág. 1 1 Simplifica las fracciones siguientes. Para ello, saca factor común cuando convenga: a) 15x 2 5x 2(x – 3) d) 9(x + 1) – 3(x + 1) 2(x + 1) a) b) 3(x – 1) 9(x – 1) 2 2 3 c) 3x 3– 9x 4 15x – 3x 2 2 e) 5x (x – 3) (x + 3) 15x (x – 3) 3 2 f ) x (3x – x 3) (3x – 1)x 3 x–3 b) x – 1 3 2 +1 c) 3x 3(1 – 3x) = 1 – 3x = –3x 2 3x (5 – x) x (5 – x) – x + 5x d) (x + 1)(9 – 3) = 6(x + 1) = 3 2(x + 1) 2(x + 1) 2 3 e) x(x – 3)(x + 3) = x (x – 9) = x – 9x 3 3 3 2 3 f) x · x (3x – 31) = x (3x – 1)3 = 1 (3x – 1)x (3x – 1)x 2 Opera y simplifica. 2 b) 3 – 2x 2 + 8x – 4x x+1 x +x 3 2 3 2 d) 5x + 15x – 10x + 215x + 2x x+3 5x a) 2 + 3 + x – 2 x 2x x c) x2 2 – 7x + 3 –9 x–3 a) 4 + 3 + 2(x – 2) = 7 + 2x – 4 = 2x + 3 2x 2x 2x 2x 2x 2 2 b) 3 – 2x + 8x – 4x = 3x – 2x + 8x – 4x · x(x + 1) = x + 1 x(x + 1) x (x + 1) x(x + 1) x(x + 1) 2 3 2 3 2 = 3x – 2x – 8x – 4x – 4x = –4x – 6x – 5x = x(x + 1) x(x + 1) 2 2 = –x (4x + 6x + 5) = –4x – 6x – 5 x(x + 1) x+1 c) 2 2 – 7x + 3 = – 7x(x + 3) + 3(x + 3)(x – 3) = (x + 3)(x – 3) x – 3 (x + 3)(x – 3) (x + 3)(x – 3) (x + 3)(x – 3) 2 2 2 = 2 – 7x – 21x + 3x – 27 = –4x – 21x – 25 (x + 3)(x – 3) (x + 3)(x – 3) 2 2 d) 5x (x + 3) – 5x (2x2+ 3) + 2x = 5x 2 – (2x + 3) + 2x = 5x 2 – 2x – 3 + 2x = 5x 2 – 3 x+3 5x Unidad 4. El lenguaje algebraico 4 Soluciones a las actividades de cada epígrafe 3 Efectúa las siguientes operaciones y simplifica. Ten en cuenta las identidades notables: 2 a) x – 1 : (x – 1) x 2 b) x (x – 2) : x – 4 x x+2 2 c) x – 2x + 1 : x – 1 x x d) 6x 2 · x –33 x e) 3x –2 3 · x (x2 + 1) x x –1 2 f ) 2x : 4x x – 1 2x – 2 5 g) x + 5 · 10 (x + 5)2 2 h) 2x · 6x3 3x 4x 2 i) 4x – 3 · 4x 8x – 6 2x 3x j) 3x –2 3 · 18(x – 1) x a) (x + 1)(x – 1) : (x – 1) = (x + 1)(x – 1) · 1 = x + 1 x x x–1 x x+2 =1 b) x(x – 2) : (x + 2)(x – 2) = x(x – 2) · x (x + 2) x (x + 2)(x – 2) 2 2 c) (x – 1) : x – 1 = (x – 1) · x = x – 1 x x x x–1 2 d) 6x (x3– 3) = 6(x – 3) = 6x – 18 x x x e) 3(x –2 1) · x(x + 1) = 3 (x + 1)(x – 1) x x 2 f ) 2x : (2x) = 2x · 2(x – 21) = 2 = 1 x – 1 2(x – 1) x – 1 (2x) 2x x 1 g) 5(x + 5) 2 = 2(x + 5) 10(x + 5) 3 h) 12x 4 = 1 x 12x 2 = 2x = x i) 4x – 3 · (2x) 2(4x – 3) 2x 2 3x = 9x 2 = 1 j) 3(x –2 1) · 18(x – 1) 18x 2x x Unidad 4. El lenguaje algebraico Pág. 2 4 Soluciones a las actividades de cada epígrafe 4 Opera y simplifica. Pág. 3 a) 6x 2 : 5x + 5x 2 2x – 3 2x + 3 4x – 9 b) x2 1– x3 + x2 – 5x 2 – 25 5 (x + 1)(5x 2 – 25) a) 6x 2 : 5x(2x + 3) + 5x(2x – 3) = (2x + 3)(2x – 3) (2x + 3)(2x – 3) ( = b) ) 2 2 6x 2 · (2x + 3)(2x – 3) = 6x = 6x 2 = 3 (2x + 3)(2x – 3) 5x(2x + 3 + 2x – 3) 5x · 4x 20x 10 5(x + 1)x 2 (x + 1)(5x 2 – 25) – 5(x 3 + x 2) – = 5 (x + 1)(5x 2 – 25) 5(x + 1)(5x 2 – 25) 5(x + 1)(5x 2 – 25) 2 2 2 2 2 2 = 5x (x + 1) – (x + 1)(5x 2– 25) – 5x (x + 1) = 5x – 5x 2+ 25 – 5x = 5(x + 1)(5x – 25) 5(5x – 25) 2 = –5x 2 + 25 = – 1 5 5(5x – 25) Unidad 4. El lenguaje algebraico
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