Giải phương trình: a) (3x – 2)(4x + 5) b) (x² – 4)(x – 2)(3 – 1x) = 0 c) (2x – 5)² – (x + 2)² = 0 d) x(2x – 9) = 3x(x – 5)

Giải phương trình:
a) (3x – 2)(4x + 5)
b) (x² – 4)(x – 2)(3 – 1x) = 0
c) (2x – 5)² – (x + 2)² = 0
d) x(2x – 9) = 3x(x – 5)

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  1. `a)` `(3x-2)(4x+5)=0`

    `<=>` \(\left[ \begin{array}{l}3x-2=0\\4x+5=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{array} \right.\)

    Vậy `x∈{2/3;-5/4}

    `b)` `(x^2-4)(x-2)(3-1x)=0`

    `<=>` `(x^2-4)(3-1x)=0`      ( ko cần lấy `x-2` vì `x^2-4` đã bao hàm nó rồi )

    `<=>` \(\left[ \begin{array}{l}x^2-4=0\\3-1x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=±2\\x=3\end{array} \right.\) 

    Vậy `x∈{2;-2;3}`

    `c)` `(2x-5)^2-(x+2)^2=0`

    `<=>` (2x-5-x-2)(2x-5-x+2)=0`

    `<=>` (x-7)(x-3)=0`

    `<=>` \(\left[ \begin{array}{l}x-7=0\\x-3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=7\\x=3\end{array} \right.\) 

    Vậy `x∈{7;3}`

    `d)` `x(2x-9)=3x(x-5)`

    `⇔2x^2-9x=3x^2-15x`

    `⇔2x^2-9x-3x^2+15x=0`

    `⇔-x^2+6x=0`

    `⇔-x(x-6)=0`

    `⇔x(x-6)=0`

    `<=>` \(\left[ \begin{array}{l}x=0\\x-6=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)

    Vậy `x∈{0;6}`

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  2. a/ Sai đề -.-

    b/ `<=>(x-2)(x+2)(x-2)(3-x)=0`

    `<=>(x+2)(3-x)(x-2)^2=0`

    `<=>S= {2;-2;3}`

     c/ `(2x-5)^2-(x+2)^2=0`

    `<=>(2x-5+x+2)(2x-5-x-2)=0`

    `<=>(3x-3)(x-7)=0`

    `<=>S={1;7}`

    d/ `x(2x-9)=3x(x-5)`

    `<=>2x^2-9x=3x^2-15x`

    `<=>3x^2-15x-2x^2+9x=0`

    `<=>x^2-6x=0`

    `<=>x(x-6)=0`

    `<=>S={0;6}`

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