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

Toán Lớp 8: Tìm a: a. 5x( x-3 ) – x + 3 = 0 b. 2x( x-2 ) + 3( x-2 ) = 0 c. ($x^{2}$ + 1 )(x-2) + 2x = 4

Toán Lớp 8: Tìm a:
a. 5x( x-3 ) – x + 3 = 0
b. 2x( x-2 ) + 3( x-2 ) = 0
c. ($x^{2}$ + 1 )(x-2) + 2x = 4

Comments ( 2 )

  1. a)
    5x(x-3)-x+3=0
    <=>5x(x-3)-(x-3)=0
    <=>(5x-1)(x-3)=0
    <=> \(\left[ \begin{array}{l}5x-1=0\\x-3=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}5x=1\\x=3\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=\dfrac{1}{5}\\x=3\end{array} \right.\) 
    vậy S={1/5;3}
    b)
    2x(x-2)+3(x-2)=0
    <=>(2x+3)(x-2)=0
    <=> \(\left[ \begin{array}{l}2x+3=0\\x-2=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}2x=-3\\x=2\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=\dfrac{-3}{2}\\x=2\end{array} \right.\) 
    vậy S={-3/2;2}
    c)
    (x^2+1)(x-2)+2x=4
    <=>(x^2+1)(x-2)+2x-4=0
    <=>(x^2+1)(x-2)+2(x-2)=0
    <=>(x^2+1+2)(x-2)=0
    <=>(x^2+3)(x-2)=0
    <=> \(\left[ \begin{array}{l}x^2+3=0\\x-2=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x^2=-3( loại : vì x^2≥0∀x) \\x=2\end{array} \right.\)  
    vậy S={2}                     

  2. Giải đáp + giải thích các bước giải:
    a)5x(x-3)-x+3=0
      ⇔5x(x-3)-(x-3)=0
        ⇔(x-3)(5x-1)=0
    ->x-3=0->x=3
    ->5x-1=0->5x=1->x=1/5
    Vậy x={1/5,3}
    b)2x(x-2)+3(x-2)=0
            ⇔(x-2)(2x+3)=0
    ->x-2=0->x=2
    ->2x+3=0->2x=-3->x=-3/2
    Vậy x={-3/2,2}
    c)(x²+1)(x-2)+2x=4
    ⇔(x²+2x+1)(x-2)+2x=4
    ⇔(x²+2x+1)(x-2)+2(x-2)=0
    ⇔(x-2)(x²+1+x-2)=0
    (x-2)(x²+3)=0
    ⇔(x-2)=0
    ⇔x=2
    Vậy x=2
     

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