a,4x(x+1)=8(x+1) b,x(x-1)-2(1-x)=0 c,2x(x-2)-(2-x)^2=0 d,(x-3)^3+3-x=0 e,5x(x-2)-(2-x)=0 f,(x+1)=(x+1)^2 g,x^2-4

1 bình luận về “a,4x(x+1)=8(x+1) b,x(x-1)-2(1-x)=0 c,2x(x-2)-(2-x)^2=0 d,(x-3)^3+3-x=0 e,5x(x-2)-(2-x)=0 f,(x+1)=(x+1)^2 g,x^2-4”

1. $a,$ $4x(x+1)=8(x+1)$

          $4x(x+1)-8(x+1)=0$

          $(x+1)(4x-8)=0$

   ⇔ $[\begin{array}{l}x+1=0\\ 4x-8=0\end{array}$ 

   ⇔$[\begin{array}{l}x=-1\\ x=2\end{array}$ 

   $Vậy$ $S=-1;2$

   $b,$ $x(x-1)-2(1-x)=0$

           $x(x-1)+2(x-1)=0$

           $(x-1)(x+2)=0$

   ⇔$[\begin{array}{l}x-1=0\\ x+2=0\end{array}$ 

   ⇔$[\begin{array}{l}x=1\\ x=-2\end{array}$ 

   $Vậy$ $S=1;-2$

   $c,$ $2x(x-2)-$$(2-x{)}^2=0$ 

          $2x(x-2)+$$(x-2{)}^2=0$

          $(x-2)(2x+x-2)=0$

          $(x-2)(3x-2)=0$

   ⇔$[\begin{array}{l}x-2=0\\ 3x-2=0\end{array}$

   ⇔$[\begin{array}{l}x=2\\ x=\frac{2}{3}\end{array}$ 

   $Vậy$……….

   $e,$ $5x(x-2)-(2-x)=0$

           $5x(x-2)+(x-2)=0$

           $(x-2)(5x+1)=0$

   ⇔$[\begin{array}{l}x-2=0\\ 5x+1=0\end{array}$ 

   ⇔$[\begin{array}{l}x=2\\ x=-\frac{1}{5}\end{array}$ 

   $Vậy$………….

   $f,$ $(x+1)=$$(x+1{)}^2$ 

         $(x+1)-$$(x+1{)}^2=0$ 

          $(x+1)[1-(x+1)]=0$

          $(x+1)(1-x-1)=0$

          $(x+1)(-x)=0$

   ⇔$[\begin{array}{l}x+1=0\\ -x=0\end{array}$ 

   ⇔$[\begin{array}{l}x=-1\\ x=0\end{array}$ 

   $Vậy$…….

   $g,$ $x^2-4x=0$ 

        $x(x-4)=0$

   ⇔$[\begin{array}{l}x=0\\ x-4=0\end{array}$ 

   ⇔$[\begin{array}{l}x=0\\ x=4\end{array}$

   $Vậy$……….

   $h,$ $(1-x{)}^2-1+x=0$ 

           $(1-x{)}^2+(1-x)=0$ 

           $(1-x)(1-x+1)=0$ 

           $(1-x)(2-x)=0$

   ⇔$[\begin{array}{l}1-x=0\\ 2-x=0\end{array}$ 

   ⇔$[\begin{array}{l}x=1\\ x=2\end{array}$ 

   $Vậy$………….

   $i,$ $x+6x^2=0$ 

          $x(1+6x)=0$

   ⇔$[\begin{array}{l}x=0\\ 1+6x=0\end{array}$ 

   ⇔$[\begin{array}{l}x=0\\ x=-\frac{1}{6}\end{array}$ 

   $Vậy$…………..

   $\text{You can't use 'macro parameter character \#' in math mode}$

             

            

    

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