Mk đg cần gấp mn giúp vs (hứa vote 5 sao) Bài 1:Tìm x, biết a) ( 2x -1) (x - 3) + 4x^2 = 1 b) x^3 - 4x^2 - x + 4 = 0 c) 3x^2 - 5x - 8 = 0

Đáp án + Giải thích các bước giải:

`a)` `(2x-1)(x-3)+4x^2=1`

`<=>2x^2-6x-x+3+4x^2=1`

`<=>6x^2-7x+3-1=0`

`<=>6x^2-7x+2=0`

`<=>6x^2-3x-4x+2=0`

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

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

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

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

Vậy `S={1/2;2/3}`

`b)` `x^3-4x^2-x+4=0`

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

`<=>(x-4)(x^2-1)=0`

`<=>(x-4)(x-1)(x+1)=0`

`<=>[(x-4=0),(x-1=0),(x+1=0):}`

`<=>[(x=4),(x=1),(x=-1):}`

Vậy `S={4;1;-1}`

`c)` `3x^2-5x-8=0`

`<=>3x^2+3x-8x-8=0`

`<=>3x(x+1)-8(x+1)=0`

`<=>(x+1)(3x-8)=0`

`<=>`\(\left[ \begin{array}{l}x+1=0\\3x-8=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=-1\\x=\dfrac{8}{3}\end{array} \right.\)

Vậy `S={-1;8/3}`.