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

Toán Lớp 8: Tìm x, biết : a. 2(x+3)-x^2-3x=0 b. x( x-3)=2x-6 c. (x-5)^2= (3+2x)^2 d. x^2- x=-2x^2+2x

Toán Lớp 8: Tìm x, biết :
a. 2(x+3)-x^2-3x=0
b. x( x-3)=2x-6
c. (x-5)^2= (3+2x)^2
d. x^2- x=-2x^2+2x

Comments ( 2 )

  1. a)2(x+3)-x²-3x=0
    ⇒2(x+3)-x(x+3)=0
    ⇒(x+3)(2-x)=0
    ⇒x+3=0⇔x=-3
    ⇒2-x=0⇔x=2
    b)x( x-3)=2x-6
    ⇒x(x-3)-2(x-3)=0
    ⇒(x-3)(x-2)=0
    ⇒x-3=0⇔x=3
    ⇒x-2=0⇔x=2

  2. Giải đáp + Lời giải và giải thích chi tiết !
    to Tìm x:
    a)
    2(x+3)-x^2-3x = 0
    <=> 2(x+3)-(x^2+3x) = 0
    <=> 2(x+3)-x(x+3) = 0
    <=> (2-x)(x+3) = 0
    <=> \(\left[ \begin{array}{l}2-x=0\\x+3=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 
    Vậy S= {2; -3}
    b)
    x(x-3) = 2x-6
    <=> x^2-3x-2x+6 = 0
    <=> (x^2-3x)-(2x-6) = 0
    <=> x(x-3)-2(x-3) = 0
    <=> (x-2)(x-3) = 0
    <=> \(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) 
    Vậy S= {2; 3}
    c)
    (x-5)^2 = (3+2x)^2
    <=> (x-5)^2-(3+2x)^2 = 0
    <=> (x-5-3-2x)(x-5+3+2x) = 0
    <=> (-x-8)(3x-2) = 0
    <=> \(\left[ \begin{array}{l}-x-8=0\\3x-2=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}-x=8\\3x=2\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=-8\\x=\dfrac{2}{3}\end{array} \right.\) 
    Vậy S= {-8; 2/3}
    Áp dụng: a^2-b^2 = (a-b)(a+b)
    d)
    x^2-x = -2x^2+2x
    <=> x^2-x+2x^2-2x = 0
    <=> (x^2-x)+(2x^2-2x) = 0
    <=> x(x-1)+2x(x-1) = 0
    <=> (x-1)(x+2x) = 0
    <=> 3x(x-1) = 0
    <=> \(\left[ \begin{array}{l}3x=0\\x-1=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\) 
    Vậy S= {0; 1}
     

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

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