Giải phương trình : a) x^2-4-(x-2)(x-5)=0 b) (x-2)(x+1)=x^2-4 c) (x+3)(x-5)+(x+3)(3x-4)=0 d) (x+6)(3x-1)+x+6=0 e) (2x-7)^2-6(2x-7)(x-3)=0

By Arianna

Giải phương trình :
a) x^2-4-(x-2)(x-5)=0
b) (x-2)(x+1)=x^2-4
c) (x+3)(x-5)+(x+3)(3x-4)=0
d) (x+6)(3x-1)+x+6=0
e) (2x-7)^2-6(2x-7)(x-3)=0

0 bình luận về “Giải phương trình : a) x^2-4-(x-2)(x-5)=0 b) (x-2)(x+1)=x^2-4 c) (x+3)(x-5)+(x+3)(3x-4)=0 d) (x+6)(3x-1)+x+6=0 e) (2x-7)^2-6(2x-7)(x-3)=0”

  1. Giải thích các bước giải:

    a, `x^2-4-(x-2)(x-5)=0`

    `⇔(x-2)(x+2)-(x-2)(x-5)=0`

    `⇔(x-2)(x+2-x+5)=0`

    `⇔(x-2).7=0`

    `⇔x=2`

    Vậy: `S={2}`

    b, `(x-2)(x+1)=x^2-4`

    `⇔(x-2)(x+1)-x^2-4=0`

    `⇔(x-2)(x+1)-(x-2)(x+2)=0`

    `⇔(x-2)(x+1-x-2)=0`

    `⇔(x-2).(-1)=0`

    `⇔x=2`

    Vậy: `S={2}`

    c, `(x+3)(x-5)+(x+3)(3x-4)=0`

    `⇔(x+3)(x-5+3x-4)=0`

    `⇔(x+3)(4x-9)=0`

    `⇔`\(\left[ \begin{array}{l}x+3=0\\4x-9=0\end{array} \right.\) 

    `⇔`\(\left[ \begin{array}{l}x=-3\\x=\dfrac{9}{4}\end{array} \right.\) 

    Vậy: `S={-3;\frac{9}{4}}`

    d, `(x+6)(3x-1)+x+6=0`

    `⇔(x+6)(3x-1+1)=0`

    `⇔(x+6).3x=0`

    `⇔`\(\left[ \begin{array}{l}x+6=0\\3x=0\end{array} \right.\) 

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

    Vậy: `S={-6;0}`

    e, `(2x-7)^2-6(2x-7)(x-3)=0`

    `⇔(2x-7)(2x-7-6x+18)=0`

    `⇔`\(\left[ \begin{array}{l}2x-7=0\\-4x+11=0\end{array} \right.\) 

    `⇔`\(\left[ \begin{array}{l}x=\dfrac{7}{4}\\x=\dfrac{11}{4}\end{array} \right.\) 

    Vậy: `S={\frac{7}{4};\frac{11}{4}}`

    Trả lời

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