Giúp mình bài tìm x này với ạ <3 a) 12x^2 - 53x + 20 = 0 b) x^3 - 5x^2 + 10x - 6 = 0 c) 1/1-x - 3x^2/x^3-1 = 2x/x^2+x+1

Đáp án:

a) $x=4$ hoặc $x=\dfrac{5}{12}$

b) $x=1$

c) $x=-\dfrac{1}{4}$

Giải thích các bước giải:

a)

$12x^2-53x+20=0\\⇔(12x^2-48x)-(5x-20)=0\\⇔12x(x-4)-5(x-4)=0\\⇔(x-4)(12x-5)=0\\⇔\left[\begin{array}{l}x-4=0\\12x-5=0\end{array}\right.\\⇔\left[\begin{array}{l}x=4\\x=\dfrac{5}{12}\end{array}\right.$

Vậy $x=4$ hoặc $x=\dfrac{5}{12}$

b)

$x^3-5x^2+10x-6=0\\⇔x^3-4x^2-x^2+6x+4x-6=0\\⇔(x^3-x^2)-(4x^2-4x)+(6x-6)=0\\⇔x^2(x-1)-4x(x-1)+6(x-1)=0\\⇔(x-1)(x^2-4x+6)=0\\⇔\left[\begin{array}{l}x-1=0\\x^2-4x+6=0\text{ (vo nghiem)}\end{array}\right.\\⇔x=1$

Vậy $x=1$

c)

$\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\,\,\,(x\ne1)\\⇔\dfrac{1}{x-1}-\dfrac{3x^2}{(x-1)(x^2+x+1)}-\dfrac{2x}{x^2+x+1}=0\\⇔\dfrac{x^2+x+1-3x^2-2x(x-1)}{(x-1)(x^2+x+1)}=0\\\to -2x^2+x+1-2x^2+2x=0\\⇔-4x^2+3x+1=0\\⇔(-4x^2+4x)-(x-1)=0\\⇔-4x(x-1)-(x-1)=0\\⇔(x-1)(-4x-1)=0\\⇔\left[\begin{array}{l}x-1=0\\-4x-1=0\end{array}\right.\\⇔\left[\begin{array}{l}x=1\text{ (loai)}\\x=-\dfrac{1}{4}\text{ (thoa man)}\end{array}\right.$

Vậy $x=-\dfrac{1}{4}$