a) 3x^2 -2x=0 b) 3x -5x +2=0 c) 2x^2 -3x -2=0 d) 6x^2 -5x +1=0 e) 6x^2 -25x -25=0 g) x^2 -4x +2=0 h) x^2 -2căn3 +2=0 i) x^2 -2x-1=0 k) x^2 -x -2 + căn

a) 3x^2 -2x=0
b) 3x -5x +2=0
c) 2x^2 -3x -2=0
d) 6x^2 -5x +1=0
e) 6x^2 -25x -25=0
g) x^2 -4x +2=0
h) x^2 -2căn3 +2=0
i) x^2 -2x-1=0
k) x^2 -x -2 + căn2=0
l) 4x^2(1+ căn3)x + căn3=0
m) 4x^2 -6x – căn2=0

0 bình luận về “a) 3x^2 -2x=0 b) 3x -5x +2=0 c) 2x^2 -3x -2=0 d) 6x^2 -5x +1=0 e) 6x^2 -25x -25=0 g) x^2 -4x +2=0 h) x^2 -2căn3 +2=0 i) x^2 -2x-1=0 k) x^2 -x -2 + căn”

  1. a) `3x^2-2x=0`

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

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

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

    Vậy `S={2;2/3}`

    b) `3x^2-5x+2=0`

    Có: `a+b+c=0 =>` \(\left[ \begin{array}{l}x=1\\x=\dfrac{2}{3}\end{array} \right.\) 

    Vậy `S={1;2/3}`

    c) `2x^2-3x-2=0`

    `<=> (2x^2-4x) + (x-2)=0`

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

    `<=>(x-2)(2x+1)=0`

    `<=>` \(\left[ \begin{array}{l}x=2\\x=\dfrac{-1}{2}\end{array} \right.\) 

    Vậy `S={2;-1/2}`

    d) `6x^2-5x+1=0`

    `<=> 6x^2-2x-3x+1=0`

    `<=> 2x(3x-1)-(3x-1)=0`

    `<=> (3x-1)(2x-1)=0`

    `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=\dfrac{1}{2}\end{array} \right.\) 

    Vậy `S={1/3 ;1/2}`

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