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

Toán Lớp 8: Tìm x. a) 9x² – 49 = 0 b) (x+3)(x² – 3x + 9) -x(x – 1)(x + 1) – 27 = 0 c)(x – 1)(x + 2) – x –

Toán Lớp 8: Tìm x. a) 9x² – 49 = 0
b) (x+3)(x² – 3x + 9) -x(x – 1)(x + 1) – 27 = 0
c)(x – 1)(x + 2) – x – 2 = 0 giúp mik vs ạ cảm ơn

Comments ( 2 )

  1. a) 9x^2 – 49 = 0
        (3x)^2 – 7^2 = 0
        ( 3x – 7 )( 3x + 7 ) = 0
    ⇔ 3x – 7 = 0
    =>  x       = 7/3
    ⇔ 3x + 7 = 0
    => x         = -7/3
    Vậy x = 7/3 hoặc x = -7/3
    b) ( x + 3 )( x^2 – 3x + 9 ) – x( x – 1 )( x + 1 ) – 27 = 0
    x^3 + 3^3 – x . x^2 + x – 27 = 0
    x^3 + 27 – 27 – x^3 + x        = 0
                                        x        = 0
    Vậy x = 0
    c) ( x – 1 )( x + 2 ) – x – 2 = 0
    ( x – 1 )( x + 2 ) – ( x + 2 ) = 0
    ( x + 2 )( x – 1 – 1 ) = 0
    ( x + 2 )( x – 2 ) = 0
    ⇔ x + 2 = 0
    => x      = -2
    ⇔ x – 2 = 0
    => x     = 2
    Vậy x = { 2; -2 }
                                       

  2. Giải đáp:
     
    Lời giải và giải thích chi tiết:
     $a,9x^{2}-49=0$
    $=>(3x)^{2}-7^{2}=0$
    $=>(3x-7)(3x+7)=0$
    =>\(\left[ \begin{array}{l}3x-7=0\\3x+7=0\end{array} \right.\)
    =>\(\left[ \begin{array}{l}3x=7\\3x=-7\end{array} \right.\)
    =>\(\left[ \begin{array}{l}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{array} \right.\) 
    $Vậy :x=±\dfrac{7}{3}$
    $b,(x+3)(x^{2}-3x+9)-x(x-1)(x+1)-27=0$
    $=>x^{3}+3^{3}-x(x^{2}-1)=27$
    $=>x^{3}+27-x^{3}+x=27$
    $=>(x^{3}-x^{3})+x=27-27$
    $=>x=0$
    $c,(x-1)(x+2)-x-2=0$
    $=>(x-1)(x+2)-(x+2)=0$
    $=>(x+2)(x-1-1)=0$
    $=>(x+2)(x-2)=0$
    =>\(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\)
    =>\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) 
    $\text{Vậy x={±2}}$
    (bài tham khảo)

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