Soluciones ecuaciones

Ecuaciones de primer grado

4. Resuelve las siguientes ecuaciones y comprueba la solución de cada una: a) 3x – 2(x + 3) = x – 3(x + 1) b) 4 + x – 4(1 – x) + 5(2 + x) = 0 c) 2x + 7 – 2(x – 1) = 3(x + 3) d) 4(2x – 7) – 3(3x + 1) = 2 – (7 – x) a) 3x – 2(x + 3) = x – 3(x + 1) → 3x – 2x – 6 = x – 3x – 3 → 3x = 3 → x = 1

Comprobación: 3 · 1 – 2(1 + 3) = 1 – 3(1 + 1) → –5 = –5

b) 4 + x – 4(1 – x) + 5(2 + x) = 0 → 4 + x – 4 + 4x + 10 + 5x = 0 → → 10x = –10 → x = –1

Comprobación: 4 – 1 – 4(1 + 1) + 5(2 – 1) = 4 – 1 – 8 + 5 = 0

c) 2x + 7 – 2(x – 1) = 3(x + 3) → 2x + 7 – 2x + 2 = 3x + 9 → 0 = 3x → x = 0

Comprobación: 2 · 0 + 7 – 2(0 – 1) = 3 · (0 + 3) → 9 = 9

d) 4(2x – 7) – 3(3x + 1) = 2 – (7 – x) → 8x – 28 – 9x – 3 = 2 – 7 + x → → –2x = 26 → x = –13

Comprobación: 4[2(–13) – 7] – 3[3(–13) + 1] = 2 – [7 – (–13)] → → –132 + 114 = 2 – 20 → –18 = –18

5. Resuelve las siguientes ecuaciones:

a) x x

53 31 2

– _{= + – b) 1 = x} x

33 – 2

+ _c) x x

5

3 4

22 + = +

d) x x x

6 5 16

128 31

– _{=– + + + e)} x x

3

2 _{4 3}

2 4

– _{= –} +

a) x x 8 x x

53 31 2 15 53 15 31 2

– _{= + – c} – _{m= c} + _{– m}

3(x – 3) = 5(x + 1) – 30 → 3x – 9 = 5x + 5 – 30 → 16 = 2x → x = 8

b) 1 = x₃+3 – x₂ → 6 · 1 = 6 xc +₃3 – x₂m → 6 = 2(x + 3) – 3x →

→ 6 = 2x + 6 – 3x → x = 0

c) x3 ₅+ = + 4 x₂2 → 2(3x – 4) = 5(x + 2) → 6x – 8 = 5x + 10 → x = 18

d) x5 ₆–16 =–x₁₂+ + + 8 x₃1 → 12 xc5 ₆–16m=12c–x₁₂+ + +8 x₃1m →

→ 2(5x – 16) = –(x + 8) + 4(x + 1) →

→ 10x – 32 = –x – 8 + 4x + 4 → 7x = 28 → x = 4

e) x2 ₃–4 3= – 4₂+ x → 6 xc2 ₃–4m=6 3c – 4+₂xm →

→ 2(2x – 4) = 18 – 3(4 + x) →

6. Resuelve y comprueba la solución de cada una de las siguientes ecuaciones:

a) x x x x

22 – 33 – –44 5–5

+ _{+ = + b)} x x x x

5

3 2

10

4 1

8

5 2

41

– – –

+ _{+ = +}

c) x x x x

55 – 245 106 604

+ _{+ = + + + d) 2x –}

2

1 (1 + 3x) – 5

3 (x – 2) = 4

1 (3 – x)

a) x₂+2 – x₃+ =3 – –x ₄4 + x –₅5 → 60 xc ₂+2 – x₃+3m=60c– –x ₄4 + x –₅5m

30(x + 2) – 20(x + 3) = –15(x – 4) + 12(x – 5) →

→ 30x + 60 – 20x – 60 = –15x + 60 + 12x – 60 → 37x = 0 → x = 0

Comprobación: 0 2+₂ – 0 3₃+ =– –0 4₄ + –₅5 → 1 – 1 = 1 – 1 → 0 = 0

b) 3x₅+2 – 4x₁₀–1 + 5x₈–2 = + x₄1 → 40 xc3 ₅+2 – 4x₁₀–1 + 5x₈–2m=40cx₄+1m

8(3x + 2) – 4(4x – 1) + 5(5x – 2) = 10(x + 1) →

→ 24x + 16 – 16x + 4 + 25x – 10 = 10x + 10 → 23x = 0 → x = 0

Comprobación: 52 – –101 + –82 = 52 + 101 – 41 = 14

c) x₅+5 – x₂₄+ = + + + 5 x₁₀6 x₆₀4 → 120 xc ₅+5 – x₂₄+5m=_{120 10}cx+ + +6 x₆₀4m

24(x + 5) – 5(x + 5) = 12(x + 6) + 2(x + 4) →

→ 24x + 120 – 5x – 25 = 12x + 72 + 2x + 8 → 5x = –15 → x = –3

Comprobación: 5–3 5+ – –3 524+ = 52 – 121 = 6019

6 3 6

60 3 4

103 601 6019

– _{+ + + =}– _{+ =}

d) 2x – ₂1 – 3₂x – 3₅x + 6₅ = ₄3 – x₄ → 20 · c2x – ₂1 – 3₂x – 3₅x + 6₅m=20·c₄3 – ₄xm →

→ 40x – 10 – 30x – 12x + 24 = 15 – 5x → 3x = 1 → _{x = 3}1

Comprobación: 2 · ₃1 ₂1 3·₂3 3·

1

53

1

5

6

– – – _{+ = 3}2 – ₂1 – ₂1 – 1₅ + 6₅ = ₃2

4 3

4 3 1

3 2

– =

7. Comprueba que las siguientes ecuaciones son de primer grado y halla sus solucio-nes:

a) (4x – 3)(4x + 3) – 4(3 – 2x)2 _{= 3x b) 2x (x + 3) + (3 – x)}2 _{= 3x (x + 1)}

c) (x x ) ( x ) x

2 1 – 2 8–1 3 4 1 – 81

2

+ ₌ +

a) (4x – 3)(4x + 3) – 4(3 – 2x)2 _{= 3x → 16x} 2 _{– 9 – 4(9 + 4x} 2 _{– 12x) = 3x →}

b) 2x(x + 3) + (3 – x)2 _{= 3x(x + 1) → 2x} 2 _{+ 6x + 9 + x} 2 _{– 6x = 3x} 2 _{+ 3x →}

→ 9 = 3x → x = 3

c) (x x₂+1) – (2x₈–1)2 = 3x₄+1 – 1₈ → 8 (ex x₂+1) – (2x₈–1)2o=8c3x₄+1 – 1₈m →

→ 4x(x – 1) – (2x – 1)2 _{= 2(3x + 1) – 1 → 4x} 2 _{– 4x – (4x} 2 _{+ 1 – 4x) = 6x + 2 – 1 →}

→ –1 = 6x + 1 8 –2 = 6x → x = –₆2 =–₃1

8. Algunas de las siguientes ecuaciones no tienen solución y otras tienen infinitas solu-ciones. Resuélvelas y comprueba los resultados.

a) 4(2x + 1) – 3(x + 3) = 5(x – 2) b) 2(x – 3) + 1 = 3(x – 1) – (2 + x)

c) x x x

2

3 _{1 2}

2 1 – –

+ = d) x x x x

4

2 _{7 2}

21

– –

+ = +

a) 8x + 4 – 3x – 9 = 5x – 10 → 5x – 5 = 5x – 10 → 0x = –5 → No tiene solución.

b) 2x – 6 + 1 = 3x – 3 – 2 – x → 2x – 5 = 2x – 5 → 0x = 0 → Tiene infinitas soluciones.

c) 2 · x2c3 +1m = 2 · xc2 – –1₂xm → 3x + 1 = 4x – 1 + x → 2 = 2x → x = 1

Comprobación: 3 1 1 2 1·₂+ = · – –1 1₂ → 2 = 2

d) 4 · cx+ 2x₄–7m=4 2·c x x+ ₂–1m → 4x + 2x – 7 = 8x + 2x – 2 → 6x – 7 = 10x – 2 →

→ – 4x = 5 → x = –₄5

Comprobación: –₄5 – 38₁₆ =–10₄ – ₈9 → –₁₆20 – 38₁₆ =–₁₆40 – ₁₆18 → –₁₆58 = –₁₆58

9. Solo una de las siguientes ecuaciones tiene solución única. Resuélvelas y comprué-balo.

a) x x

21 2 2 4–3

+ = + b) x x x x

12

4 3

4

2 1

31 3 6 1

– – + = – – +

c) x x x

3 1

53 1526 42

– –

+ _{+ = + d)} (x ) x (x ) x

161 – 12 16–1 – 24

2 2

+ _{+ =} +

a) 4 · cx+₂1m=4 2·c + 2x₄–3m → 2x + 2 = 8 + 2x – 3 → 2x + 2 = 2x + 5 → 0x = 3 →

→ No tiene solución.

b) 12 · c4x₁₂– –3 2x₄+1m=12·cx₃– –1 3x₆+1m → 4x – 3 – 6x – 3 = 4x – 4 – 6x – 2 →

→ –2x – 6 = –2x – 6 → 0x = 0 → Tiene infinitas soluciones.

c) 30 · c1+₃x – x+₅3m=30·c₁₅26 – 4+₂xm → 10 + 10x – 6x – 18 = 52 – 60 – 15x →

→ –8 + 4x = –8 – 15x → 19x = 0 → x = 0

d) 16 · (= x₁₆+1)2 – 1+₂xG=16· (= x₁₆–1)2 – 2+₄xG → (x + 1)2 _{– 8 – 8x = (x – 1)}2 _{– 8 – 4x →}

→ x 2 + 2x + 1 – 8 – 8x = x 2 – 2x + 1 – 8 – 4x → x 2 – 6x – 7 = x 2 – 6x – 7 → → 0x = 0 → Tiene infinitas soluciones.

10. Resuelve.

a) (x ) (x ) x x

3

2 ₃

5

1 ₅

5 3

3 2

15 4

– + – = d + n+ b) 2x –

2

1 (1 + 3x) = 5

3 (x – 2) + 4

1 (3 – x)

c) ( x) ( x ) x x

3 4 2

4

3 2 1 4 7 2 1

4 3

– – – = – d – n– d) x (8x – 1) – (3x – 4)2 _{= x (7 – x) – 2(x – 4)}

a) x2₃ – ₃6 + x₅ – ₅5 = 3₅x + ₁₅6 + 4₁₅x → 15 · c2₃x – ₃6 + ₅x – 5₅m=15·c3₅x + ₁₅6 + ₁₅4xm →

→ 10x – 30 + 3x – 15 = 9x + 6 + 4x → 13x – 45 = 13x + 6 →

→ 0x = 51 → No tiene solución.

b) 20 · c2x– ₂1 – 3₂xm=20·c3₅x – 6₅ + 3₄ – ₄xm → 40x – 10 – 30x = 12x – 24 + 15 – 5x →

→ 10x – 10 = 7x – 9 → 3x = 1 → _{x = 3}1

c) ₃8 – 4₃x – 6₄x _{+ = 4x – 7x + 24}3 7 – ₄3 →

→ 12 · c₃8 – 4₃x – 6₄x + 3₄m=12 4·c x –7x+ ₂7 – 3₄m →

→ 32 – 16x – 18x + 9 = 48x – 84x + 42 – 9 → 41 – 34x = 33 – 36x → 2x = –8 →

→ _{x = – 2}8 = – 4

d) 8x 2 _{– x – (9x} 2 _{– 24x + 16) = 7x – x} 2 _{– 2x + 8 → –x} 2 _{+ 23x – 16 = –x} 2 _{+ 5x + 8 →}

→ 18x = 24 → _{x = 18}24 3 4 =

Página 116

Ecuaciones de segundo grado

11. Resuelve las siguientes ecuaciones de segundo grado sin utilizar la fórmula de reso-lución:

a) 3x 2 _{– 12x = 0 b) x – 3x} 2 _{= 0}

c) 2x 2 _{– 5x = 0 d) 2x} 2 _{– 8 = 0}

e) 9x 2 _{– 25 = 0 f ) 4x} 2 _{+ 100 = 0}

g) 16x 2 _{= 100 h) 3x} 2 _{– 6 = 0}

a) 3x 2 _{– 12x = 0 → 3x(x – 4) = 0 x}

x==04

b) x – 3x 2 _{= 0 → x(1 – 3x) = 0}

/

x x

0 1 3 = =

c) 2x 2 _{– 5x = 0 → x(2x – 5) = 0}

/

x x==05 2

d) 2x 2 – 8 = 0 → 2x 2 = 8 → x 2 = 4 x

x

2 2 – = =

e) 9x 2 _{– 25 = 0 → 9x} 2 _{= 25 → x} 2

= 925 xx==5 3–/5 3/ f) 4x 2 _{+ 100 = 0 → 4x} 2 _{= –100 No tiene solución.}

g) 16x 2 _{= 100 → x} 2 ₌

16

100 / /

/ /

x

x==10–104 54==–25 2

h) 3x 2 – 6 = 0 → 3x 2 = 6 → x 2 = 2 x

x

2 2 – = =

12. Resuelve.

a) x 2 _{+ 4x – 21 = 0 b) x} 2 _{+ 9x + 20 = 0}

c) 9x 2 _{– 12x + 4 = 0 d) x} 2 _{+ x + 3 = 0}

e) 4x 2 _{+ 28x + 49 = 0 f ) x} 2 _{– 2x + 3 = 0}

g) 4x 2 _{– 20x + 25 = 0 h) –2x} 2 _{+ 3x + 2 = 0}

a) x 2 + 4x – 21 = 0 → x = –4± 16 21 4₂+ · = –4 10₂± x

x

3 7 – = =

b) x 2 _{+ 9x + 20 = 0 → x =} ± · ±

2

9 81 4 20

2 9 1

– – ₌ – _x

x==––54

c) 9x 2 _{– 12x + 4 = 0 → x =} ± · · ±

18

12 144 4 9 4

18 12 0

3 2

– _{= =}

d) x 2 _{+ x + 3 = 0 → x =} ± ·

2

1 1 4 3

e) 4x 2 + 28x + 49 = 0 → x = –28± 784 4 4 49₈– · · = –28 0₈± =–₂7

f) x 2 _{– 2x + 3 = 0 → x = ±} ·

2

2 4 4 3– _{No tiene solución.}

g) 4x 2 – 20x + 25 = 0 → x = 20± 400 4 4₈– · ·25 = 20 0₈± = ₂5

h) –2x 2 _{+ 3x + 2 = 0 → x =} ± ( ) · ±

4

3 9 4 2 2

4

3 5

–

– – –

– –

= x / /

x==2–2 4=–1 2

13. Resuelve igualando a cero cada factor:

a) x (3x – 1) = 0 b) 3x (x + 2) = 0 c) (x + 1)(x + 3) = 0 d) (x – 5)(x + 5) = 0 e) (x – 5)2 = 0 f ) (2x – 5)2 = 0 a) x = 0; 3x – 1 = 0 → _{x = 3}1 Soluciones: x = 0; _{x = 3}1

b) 3x = 0; x + 2 = 0 → x = –2 Soluciones: x = 0; x = –2

c) x + 1 = 0; x + 3 = 0 Soluciones: x = –1; x = –3

d) x – 5 = 0; x + 5 = 0 Soluciones: x = 5; x = –5

e) x – 5 = 0 Solución: x = 5

f) 2x – 5 = 0 _{Solución: x = 2}5

14. Opera y resuelve.

a) (x – 2)(3x + 2) = (x – 4)(2x + 1) b) (x – 1)2 _{+ (1 – x)(x + 2) = 0}

c) (x + 1)2 = (x + 1)(2x – 3) d) 5(x + 2)2 – (7x + 3)(x + 2) = 0 a) 3x 2 _{+ 2x – 6x – 4 = 2x} 2 _{+ x – 8x – 4 → 3x} 2 _{– 4x – 4 = 2x} 2 _{– 7x – 4 →}

→ x 2 _{+ 3x = 0 → x · (x + 3) = 0 → x}

1 = 0; x2 = –3

b) x 2 _{– 2x + 1 + x + 2 – x} 2 _{– 2x = 0 → x + 3 = 0 → x = –3}

c) x 2 _{+ 2x + 1 = 2x} 2 _{– 3x + 2x – 3 → x} 2 _{+ 2x + 1 = 2x} 2 _{– x – 3 → –x} 2 _{+ 3x + 4 = 0 →}

→ x = – ±3 32– · – ·_–24 ( )1 4 = – ±3 _–29 16+ = – ±3_–225 = –3 5_–±₂ → x1 = –1; x2 = 4

d) 5 · (x 2 + 4x + 4) – (7x 2 + 14x + 3x + 6) = 0 → 5x 2 + 20x + 20 – 7x 2 – 14x – 3x – 6 = 0

→ –2x 2 _{+ 3x + 14 = 0 → x =}

· ( )

± · ( ) · ± ±

2 2

3 3 4 2 14

4

3 121

4 3 11 –

– – –

– –

– –

2

= = →

→ x₁ = –2; x₂ = 14₄ = ₂7

15. Resuelve las siguientes ecuaciones:

a) (2x + 1)(x – 3) = (x + 1)(x – 1) – 8 b) (2x – 3)(2x + 3) – x (x + 1) – 5 = 0

c) (2x + 1)2 _{= 4 + (x + 2)(x – 2) d) (x + 4)}2 _{– (2x – 1)}2 _{= 8x}

a) (2x + 1)(x – 3) = (x + 1)(x – 1) – 8 → 2x 2 _{– 6x + x – 3 = x} 2 _{– 1 – 8 →}

→ x 2 _{– 5x + 6 = 0 → x = ±} · _{8 x} ±

2

5 25 4 6

2

5 1

– _{= x}

b) (2x – 3)(2x + 3) – x(x + 1) – 5 = 0 → 4x 2 _{– 9 – x} 2 _{– x – 5 = 0 → 3x} 2 _{– x – 14 = 0 →}

→ x = ±1 1 4 3– ₆· · (–14) = 1±₆169 = 1 13±₆ x /

x==7 3–2

c) (2x + 1)2 _{= 4 + (x + 2)(x – 2) → 4x} 2 _{+ 1 + 4x = 4 + x} 2 _{– 4 → 3x} 2 _{+ 4x + 1 = 0 →}

→ x = –4± 16 4 3 1₆– · · = –4₆± 4 = –4 2₆± x /

x==––1 31

d) (x + 4)2 _{– (2x – 1)}2 _{= 8x → x} 2 _{+ 16 + 8x – (4x} 2 _{+ 1 – 4x) – 8x = 0 →}

→ x 2 _{+ 16 + 8x – 4x} 2 _{– 1 + 4x – 8x = 0 → –3x} 2 _{+ 4x + 15 = 0 →}

→ x = –4± 16 4–_–6· ( ) ·–3 15 = –4±_–6196 = –4 14_–±₆ x /

x==3–53

16. Resuelve las ecuaciones siguientes:

a) ( x )( x ) ( x ) 4

5 4 5 4

2

3 1 9

– ₊ – 2–

= b) x x( ) x x( ) x

3 – –1 4 + +1 312+ =4 0 c) (x )(x ) (x )(x ) x

12

1 2

6

1 2 ₁

33

– + _– + – _{– =} – _{d) (}x ) x _x

15

1 3 1

51 0 – 2–

+ + + =

e) x (x ) x (x )

21 – –41 – 32 –62 61

2 2

+ _{+ + =}

a) (5x–4 5)(₄ x+4) = (3x –1₂)2–9 → 25x2₄–16 = 2 9( x2+1 6₄– x –9) → → 25x 2 _{– 16 = 18x} 2 _{+ 2 – 12x – 18 → 7x} 2 _{+ 12x = 0 →}

→ x(7x + 12) = 0

/

x

x==0–12 7

b) x x₃( – –1) ₄x x( + +1) 3x₁₂+ = 0 4 → 12cx x₃( – –1) ₄x x( + +1) 3x₁₂+4m →

→ 4x(x – 1) – 3x(x + 1) + 3x + 4 = 0 → 4x 2 _{– 4x – 3x} 2 _{– 3x + 3x + 4 = 0 →}

→ x 2 _{– 4x + 4 = 0 → x = ±} ·

2

4 16 4 4– _{= 2}

c) (x–1₁₂)(x+2) – (x+1)(₆x –2) –1= x–₃3 → x2+₁₂x – – – – –2 x2 ₆x 2 1= x₃–3 →

→ 12 xe 2+₁₂x – – – – –2 x2 ₆x 2 1 12o= cx ₃–3m →

→ x 2 _{+ x – 2 – 2(x} 2 _{– x – 2) – 12 = 4(x – 3) →}

→ x 2 _{+ x – 2 – 2x} 2 _{+ 2x + 4 – 12 = 4x – 12 → –x} 2 _{– x + 2 = 0 →}

→ x 2 _{+ x – 2 = 0 → x =} ± ( ) ±

2

1 1 4 2

2 1 3

– – – ₌ – _x

x==1–2

d) (x–1)2₁₅–3x+ + + = 0 1 x₅1 → 15=(x –1)₁₅2–3x+ + +1 x₅1G = 0 →

→ x 2 _{– 2x + 1 – 3x + 1 + 3x + 3 = 0 → x} 2 _{– 2x + 5 = 0 →}

e) x₂+1 – (x –₄1)2 – x+ +₃2 (x–₆2)2 = 1₆ →

→ 12ex₂+1 – (x –₄1)2 – x+ +₃2 (x –₆2)2o=12· 1₆ →

→ 6(x + 1) – 3(x 2 – 2x + 1) – 4(x + 2) + 2(x 2 – 4x + 4) = 2 → → 6x + 6 – 3x 2 _{+ 6x – 3 – 4x – 8 + 2x} 2 _{– 8x + 8 = 2 →}

→ –x 2 _{+ 3 = 0 → x} 2 _{= 3 x}

x==–33

17. Resuelve.

a) (x ) x x x

8

7 5 ₂

2 9

4 11

– _{+ – =d – nd – n b)} x ( x) 33 – 4–9 31

2

+ ₌

c) ( x )( x ) x x x

21

3 1 2 3

7 3 3 –2

2 2

+ + _{+ + = + d) x} ( x ) x

3–4 2 8–2 7 12–10

2 2 2

+ =

a) x 87 –35 + x – 2 = x 2 _{– x} x

4 11

2 9

8 99

– + →

→ 8 · c7x –₈35 +x –2 8m= ·cx2– 11₄x – 9₂x + 99₈ m →

→ 7x – 35 + 8x – 16 = 8x 2 _{– 22x – 36x + 99 → 15x – 51 = 8x} 2 _{– 58x + 99 →}

→ 8x 2 _{– 73x + 150 = 0}

x = 73± (–73_{2 8})2_·–4 8· ·150 = 73± 5329 4 800₁₆ – = 73±₁₆529 = 73 23₁₆± →

→ x₁ = 6; x₂ = ₁₆50 = 25₈

b) 9 · ex+₃3 – (4–₉x)2o=_{9 3}·c1m → 3x + 9 – (4 – x)2 _{= 3 →}

→ 3x + 9 – 16 + 8x – x 2 _{= 3 → –x} 2 _{+ 11x – 10 = 0 →}

→ x = –11± 112–_–4₂· ( ) · (–1 –10) = –11_–±₂ 81 = –11 9_–2± → x₁ = 1; x₂ = 10

c) 21 · (= 3x+1 2₂₁)( x+3) + x2₇+3G=21·ex2+₃x–2o →

→ (3x + 1) · (2x + 3) + 3x 2 + 9 = 7x 2 + 7x – 14 →

→ 6x 2 _{+ 9x + 2x + 3 + 3x} 2 _{+ 9 = 7x} 2 _{+ 7x – 14 → 9x} 2 _{+ 11x + 12 = 7x} 2 _{+ 7x – 14 →}

→ 2x 2 _{+ 4x + 26 = 0 → x} 2 _{+ 2x + 13 = 0 → x =} ±

2

2 2 4 1 13

2

2 48

– ± 2– · · – –

= →

→ No tiene solución.

d) 24 · =x2₃–4 + (2x ₈–2)2G=24·e7x2₁₂–10o → 8x 2 _{– 32 + 3 · (2x – 2)}2 _{= 14x} 2 _{– 20 →}

→ 8x 2 _{– 32 + 12x} 2 _{– 24x + 12 = 14x} 2 _{– 20 → 20x} 2 _{– 24x – 20 = 14x} 2 _{– 20 →}

→ 6x 2 _{– 24x = 0 → 6x(x – 4) = 0 → x}

18. Resuelve las siguientes ecuaciones:

a) x

x xx

5 – 3 = + b) 1 x

x xx xx 32 – 1 –3 42–

2

+ _{= +}

c) x

x xx xx 23 – 1 –3 42–

2

+ _{= + d)}

x x x

15 2

72 6– ₂

2

= +

a) x · c5x – 3_xm=x·cx_x+1m → 5x 2 _{– 3 = x + 1 → 5x} 2 _{– x – 4 = 0 →}

→ x = – –( ) ± ( )1 –1_{2 5·}2–4 5· · ( )–4 = 1±₁₀81 = 1 9₁₀± → x₁ = 1; x_{2 = – 10}8 =–₅4

b) 6x · cx₃+2 – 1_xm=6x·cx –_x 3 + 4₂–_xx2m → 2x 2 + 4x – 6 = 6x – 18 + 12 – 3x 2 →

→ 5x 2 _{– 2x = 0 → x · (5x – 2) = 0 → x}

1 = 0; x2 = ₅2

Debemos descartar la solución x₁ = 0, ya que anula algunos denominadores.

c) 2x xc ₂+3 – 1_xm=2xcx _x–3 + 4–_2xx2m → x 2 _{+ 3x – 2 = 2x – 6 + 4 – x} 2 _→

→ 2x 2 _{+ x = 0 → x(2x + 1) = 0 → x}

1 = 0; x2 = 2–1

Debemos descartar la solución x₁ = 0, ya que anula algunos denominadores.

d) 2x 2

x x _x x

15 ₂

2

72 6– ₂

2

2

= +

c m e o → 30x = 72 – 6x + 4x 2 _{→ 4x} 2 _{– 36x + 72 = 0 →}

→ x 2 _{– 9x + 18 = 0}

x = ( ) ± ( )– –9 –₂9 2–4 18· = 9± 81 72₂ – = 9±₂ 9 = 9 3₂± → x₁ = 6; x₂ = 3

Aplica lo aprendido

19. La suma de tres números naturales consecutivos es igual al quíntuple del menor menos 11. ¿Cuáles son esos números?

Llamemos x, x + 1, x + 2 a los números. Así:

x + x + 1 + x + 2 = 5x – 11 → 14 = 2x → x = 7

Los números son 7, 8 y 9.

20. Calcula un número tal que sumándole su mitad se obtiene lo mismo que restando 6 a los 9/5 de ese número.

x + x₂ = x – 6 ₅9 → 10 x xb + ₂l=10c₅9x–6m → 10x + 5x = 18x – 60 →

→ 60 = 3x → x = 20