Tìm x, biết a) x(2x-7)-4x+14=0 b)x(x-1)+2x-2=0 c)x+x^2-x^3-x^4=0 d)2x^3+3x^2+2x+3=0

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1. Đáp án:

    

   Giải thích các bước giải:

   a/ $x(2x-7)-4x+14=0$
   ⇔ $2x^2-7x-4x+14=0$
   ⇔ $2x^2-11x+14=0$
   ⇔ $2x^2-4x-7x+14=0$
   ⇔ $2x(x-2)-7(x-2)=0$
   ⇔ $(x-2)(2x-7)=0$
   ⇔\(\left[ \begin{array}{l}x=2\\x=\frac{7}{2}\end{array} \right.\)

   b/ $x(x-1)+2x-2=0$
   ⇔ $x^2-x+2x-2=0$
   ⇔ $x^2+x-2=0$
   ⇔ $x^2+2x-x-2=0$
   ⇔ $(x+2)(x-1)=0$
   ⇔ \(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)

   c/ $x+x^2-x^3-x^4=0$
   ⇔ $x(1+x)-x^3(1+x)=0$
   ⇔ $x(x+1)(1-x^2)=0$
   ⇔ $x(x+1)(1-x)(x+1)=0$
   ⇔ $x(1-x)(x+1)^2=0$
   ⇔ \(\left[ \begin{array}{l}x=0\\x=±1\end{array} \right.\)

   d/ $2x^3+3x^2+2x+3=0$
   ⇔ $2x(x^2+1)+3(x^2+1)=0$
   ⇔ $(x^2+1)(2x+3)=0$
   Vì $x^2+1 > 0$ nên $2x+3=0$
   ⇔ $2x=-3$
   ⇔ $x=-\frac{3}{2}$
   Chúc bạn học tốt !!!

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2. `a. x(2x-7)-4x+14=0`

   `=>2x^2-7x-4x+14=0`

   `=>2x^2-4x  -7x+14=0`

   `=> 2x(x-2)-7(x-2)=0`

   `=>(x-2)(2x-7)=0`

   `=>`\(\left[ \begin{array}{l}x=2\\x=7/2\end{array} \right.\) 

   `b. x(x-1)+2x-2=0`

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

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

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

   `c.x+x^2-x^3-x^4=0`

   `=>x(x^3+x^2-x-1)=0`

   `=>x[x^2(x+1)-(x-1)]=0`

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

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

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

   `d.2x^3+3x^2+2x+3=0`

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

   `=>(2x+3)(x^2+1)=0`

   `=>2x+3=0(do    x^2+1\ne 0)`

   `=>x=\frac{-3}{2}`

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