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

Toán Lớp 8: Tìm x, biết a) (x+2)^2-x(x+3)+5x=-20 b) 5x^3-10x^2+5x=0 c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0 d) (x+1)^3-(x-1)^3-6(x-1)^2=-19

Toán Lớp 8: Tìm x, biết
a) (x+2)^2-x(x+3)+5x=-20
b) 5x^3-10x^2+5x=0
c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0
d) (x+1)^3-(x-1)^3-6(x-1)^2=-19

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết :
    a) ( x + 2 )^2 – x ( x + 3 ) + 5x = – 20
    ⇔ x^2 + 4x + 4 – x^2 – 3x + 5x = – 20
    ⇔ 6x + 4 = – 20
    ⇔ 6x = – 24
    ⇔ x = – 4
    Vậy x = – 4
    b) 5x^3 – 10x^2 + 5x = 0
    ⇔ 5x ( x^2 – 2x + 1 ) = 0
    ⇔ 5x ( x – 1 )^2 = 0
    ⇔ x ( x – 1 )^2 = 0
    ⇔ \(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\) 
    Vậy x = 0 ; x = 1
    c) ( x^2 – 1 )^3 – ( x^4 + x^2 + 1 ) ( x^2 – 1 ) = 0
    ⇔ ( x^2 – 1 ) . [ ( x^2 – 1 )^2 – ( x^4 + x^2 + 1 ) ] = 0
    ⇔ ( x^2 – 1 ). ( x^4 – 2x^2 + 1 – x^4 – x^2 – 1 ) = 0
    ⇔ ( x^2 – 1 ) ( – 3x^2 ) = 0
    ⇔ \(\left[ \begin{array}{l}x^2-1=0\\-3x^2 = 0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=1\\x=-1\\x=0\end{array} \right.\) 
    Vậy x = – 1 ; x = 1 ; x = 0
    d) ( x + 1 )^3 – ( x – 1 )^3 – 6 ( x – 1 )^2 = – 19
    ⇔ x^3 + 3x^2 + 3x + 1 – x^3 + 3x^2 – 3x + 1 – 6x^2 + 12x – 6 = – 19
    ⇔ – 4 + 12x = – 19
    ⇔ 12x = – 15
    ⇔ x = – 5/4
    Vậy x = – 5/4

  2. Giải đáp + Lời giải và giải thích chi tiết:
    a) (x + 2)^2 – x(x + 3) + 5x = -20
    →x^2 + 2.x.2 + 2^2 – x.x – x.3 + 5x + 20 = 0
    →x^2 + 4x + 4 – x^2 – 3x + 5x + 20 = 0
    →(x^2 – x^2) + (4x – 3x + 5x) + (4 + 20) = 0
    →6x + 24 = 0
    →6x          = 0 – 24
    →6x          = -24
    →x            = -24 : 6
    →x            = -4
    Vậy x = -4
    b) 5x^3 – 10x^2 + 5x = 0
    ↔5x(x^2 – 2x + 1) = 0
    ↔5x.(x – 1)^2 = 0
    -> \(\left[ \begin{array}{l}5x = 0\\(x – 1)^2 = 0\end{array} \right.\) 
    -> \(\left[ \begin{array}{l}x = 0 : 5 \\(x – 1)^2 = 0^2\end{array} \right.\) 
    -> \(\left[ \begin{array}{l}x=0\\x – 1 = 0\end{array} \right.\) 
    -> \(\left[ \begin{array}{l}x=0\\x=0 + 1\end{array} \right.\) 
    -> \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\) 
    Vậy x = 0 hoặc x = 1
    c) (x^2 – 1)^3 – (x^4 + x^2 + 1)(x^2 – 1) = 0
    ↔ (x^2 – 1)[(x^2 – 1)^2 – (x^4 + x^2 + 1)] = 0
    ↔ (x – 1)(x + 1)[(x^4 – 2x^2 + 1) – x^4 – x^2 – 1] = 0
    ↔ (x – 1)(x + 1)[x^4 – 2x^2 + 1 – x^4 – x^2 – 1] = 0
    ↔ (x – 1)(x + 1)[(x^4 – x^4) + (-2x^2 – x^2) + (1 – 1)] = 0
    ↔ (x – 1)(x +1)[-3x^2] = 0
    ↔ (x – 1)(x + 1)(-3x^2) = 0
    ↔ 3x^2(x – 1)(x + 1) = 0
    →\(\left[ \begin{array}{l}3x^2 = 0\\x – 1 = 0\\x + 1 = 0\end{array} \right.\) 
    →\(\left[ \begin{array}{l}x^2 = 0 : 3\\x=0 + 1\\x = 0 – 1\end{array} \right.\) 
    ->\(\left[ \begin{array}{l}x^2 = 0\\x=1\\x = -1\end{array} \right.\) 
    -> \(\left[ \begin{array}{l}x=0\\x=1\\x = -1\end{array} \right.\) 
    Vậy x = 0 hoặc x = 1 hoặc x = -1
    d) (x+1)^3-(x-1)^3-6(x-1)^2=-19
    →(x^3 + 3x^2 . 1 + 3x.1^2 + 1^3) – (x^3 – 3x^2 . 1 + 3x.1^2 – 1^3) – 6(x^2 – 2x.1 + 1^2) + 19 = 0
    →(x^3 + 3x^2 + 3x + 1) – (x^3 – 3x^2 + 3x – 1) – 6(x^2 – 2x + 1) + 19 = 0
    →x^3 + 3x^2 + 3x + 1 – x^3 + 3x^2 – 3x + 1 – 6x^2 + 12x – 6 + 19 = 0
    →(x^3 – x^3) + (3x^2 + 3x^2 – 6x^2) + (3x – 3x + 12x) + (1 + 1 – 6 + 19) = 0
    →12x + 15 = 0
    →12x          = 0 – 15
    →12x          = -15
    →x              = -15 : 12
    →x              = -5/4
    Vậy x = -5/4

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

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