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

Toán Lớp 7: (2+3x)(2-3x) -x(9x+1) =0 (x-2)^2 -3x(x-2) =0

Toán Lớp 7: (2+3x)(2-3x) -x(9x+1) =0
(x-2)^2 -3x(x-2) =0

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    a) $(2+3x)(2-3x) – x(9x+1)=0$
    ⇔$4 -9x^2-9x^2-x = 0$
    ⇔$18x^2+x -4    =0$
    ⇔$(9x -4)(2x+1)  =0$
    ⇔\(\left[ \begin{array}{l}9x-4=0\\2x+1=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=\frac{4}{9}\\x=\frac{-1}{2}\end{array} \right.\) 
    Vậy $S={\frac{4}{9};\frac{-1}{2}}$
    b) $(x-2)^2 -3x(x-2) =0$
    ⇔$x^2 -4x +4 -3x^2+6x =0$
    ⇔$2x^2 -2x -4           =0$
    ⇔$2(x^2 -x -2)       =0$
    ⇔$2(x-2)(x+1) =0$
    ⇔\(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\) 
    VẬy $S={2;-1}$
    mk xin hay nhất ạ, mk rất cần ạ 

  2. Lời giải và giải thích chi tiết:
    a, (2+3x)(2-3x)-x(9x+1)=0
    <=>[2^2-(3x)^2]-x(9x+1)=0
    <=>4-9x^2-9x^2-x=0
    <=>18x^2+x-4=0
    <=>(9x-4)(2x+1)=0
    <=>\(\left[ \begin{array}{l}9x-4=0\\2x+1=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=\dfrac{4}{9} \\x=\dfrac{-1}{2}\end{array} \right.\)
    Vậy S={4/9;-1/2}
    b, (x-2)^2-3x(x-2)=0
    <=>(x-2)(x-2-3x)=0
    <=>(x-2)(-2x-2)=0
    <=>\(\left[ \begin{array}{l}x-2=0\\-2x-2=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
    Vậy S={2;-1}

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