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

Toán Lớp 8: tìm x a) 2x(x+2)-(x-2)(x+2) = 3 b)9(3x-2)= x(2-3x)

Toán Lớp 8: tìm x a) 2x(x+2)-(x-2)(x+2) = 3
b)9(3x-2)= x(2-3x)

Comments ( 2 )

  1. $a)$ $2x(x+2) – (x-2)(x+2) = 3$
    $(x+2).[2x-(x-2)] = 3$
    $(x+2).(2x-x+2) = 3$
    $(x+2).(x+2) = 3$
    $(x+2)^{2} = 3$
    $x+2 =$ $\sqrt[]{3}$ $\text{hoặc}$ $x+2 =$ $-\sqrt[]{3}$
    $x=$ $\sqrt[]{3}-2$ $\text{hoặc}$ $x =$ $-\sqrt[]{3}-2$
    $b)$ $9(3x-2)= x(2-3x)$
    $27x-18= 2x-3x^{2}$
    $27x-18 – 2x+3x^{2} = 0$
    $(3x^{2} – 2x) + (27x – 18) = 0$
    $x.(3x – 2) + 9.(3x – 2) = 0$
    $(3x – 2)(x + 9) = 0$
    $3x – 2 = 0$ $\text{hoặc}$ $x + 9 = 0$
    $x =$ $\dfrac{2}{3}$ $\text{hoặc}$ $x = -9$

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

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

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