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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Bài 7: : Tìm x biết: a) (x – 3)(x2 + 3x + 9) + x(x + 2)(2 – x) = 1; b) (x + 1)^3- (x – 1)^3- 6(x – 1)^2 = -10; c) (x+2)(x^2-2x +4)

Toán Lớp 8: Bài 7: : Tìm x biết:
a) (x – 3)(x2 + 3x + 9) + x(x + 2)(2 – x) = 1;
b) (x + 1)^3- (x – 1)^3- 6(x – 1)^2 = -10;
c) (x+2)(x^2-2x +4) – x(x^2 +2) = 0

Comments ( 2 )

  1. a) $ ( x – 3 ) ( x^{2} + 3x + 9 ) + x ( x + 2 ) ( 2 – x ) = 1 $
    $ ⇒ x^{3} – 3^{3} + x ( 4 – x^{2} ) = 1 $
    $ ⇒ x^{3} – 27 + 4x – x^{3} = 1 $
    $ ⇒ 4x = 28 $
    $ ⇒ x = 7 $
    Vậy $ x = 7 $
    b) $ ( x + 1 )^{3} – ( x – 1 )^{3} – 6 ( x – 1 )^{2} = -10 $
    $ ⇒ x^{3} + 3x^{2} + 3x + 1 – ( x^{3} – 3x^{2} + 3x – 1 ) – 6 ( x^{2} – 2x + 1 ) = – 10 $
    $ ⇒ x^{3} + 3x^{2} + 3x + 1 – x^{3} + 3x^{2} – 3x + 1 – 6x^{2} + 12x – 6 = – 10 $
    $ ⇒ 12x = – 6 $
    $ ⇒ x = \dfrac{-1}{2} $
    Vậy $  x = \dfrac{-1}{2} $
    c) $ ( x + 2 ) ( x^{2} – 2x + 4 ) – x ( x^{2} + 2 ) = 0 $
    $ ⇒ x^{3} + 8 – x^{3} – 2x = 0 $
    $ ⇒ -2x = -8 $
    $ ⇒ x = 4 $
    Vậy $ x = 4 $
     

  2. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    a) (x – 3)(x² + 3x + 9) + x(x + 2)(2 – x) = 1
    ⇔ (x – 3)(x² + 3x + 3²) – x(x + 2)(x – 2) = 1
    ⇔ x³ – 3³ – x(x² – 4) = 1
    ⇔ x³ – 27 – x³ + 4x = 1
    ⇔ 4x = 28
    ⇔ x = 7
    b) (x + 1)³ – (x – 1)³ – 6(x – 1)² = -10
    ⇔ (x³ + 3x² + 3x + 1) – (x³ – 3x² + 3x – 1) – 6(x² – 2x + 1) = -10
    ⇔ x³ + 3x² + 3x + 1 – x³ + 3x² – 3x + 1 – 6x² + 12x – 6 = -10
    ⇔ 12x – 4 = -10
    ⇔ 12x = -6
    ⇔ x = $\frac{-1}{2}$ 
    c) (x + 2)(x² – 2x + 4) – x(x² + 2) = 0
    ⇔ (x + 2)(x² – 2x + 2²) – x³ – 2x = 0
    ⇔ x³ + 2³ – x³ – 2x = 0
    ⇔ 8 – 2x = 0
    ⇔ 2x = 8
    ⇔ x = 4

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222-9+11+12:2*14+14 = ? ( )