tìm x biết 1.x(x+1)-x^2+1=0 2.4x(x-2)-6+3x=0 3.x(x+2)-3(x+2)=0 4.8x(x-5)-2x+10=0 5.x(2x-3)-2(3-2x)=0

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1. Đáp án:

    

   Giải thích các bước giải:

    1) $x(x+2)-x^{2}+1=0_{}$ 

   ⇔ $x^{2}-x-^{2}+1=0_{}$ 

   ⇔ $x+1=0_{}$ 

   ⇔ $x=-1_{}$ 

   2) $4x(x-1)-3(x+2)=0_{}$ 

   ⇔ $4x(x-2)+3x-6=0_{}$ 

   ⇔ $4x(x-2)+3(x-2)=0_{}$ 

   ⇔ $(x-2)(4x+3)=0_{}$ 

   ⇔ \(\left[ \begin{array}{l}x-2=0\\4x+3=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=2\\x=-\frac{3}{4}\end{array} \right.\) 

   3) $x(x+2)-3(x+2)=0_{}$ 

   ⇔ $(x-3)(x+2)=0_{}$ 

   ⇔ \(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\) 

   4) $8x(x-5)-2x+10=0_{}$ 

   ⇔ $8x(x-5)-2(x-5)=0_{}$ 

   ⇔ $(8x-2)(x-5)=0_{}$ 

   ⇔ \(\left[ \begin{array}{l}8x-2=0\\x-5=0\end{array} \right.\) ⇔  \(\left[ \begin{array}{l}x=\frac{1}{4}\\x=5\end{array} \right.\)

   5) $x(2x-3)-2(3-2x)=0_{}$ 

   ⇔ $x(2x-3)+2(2x-3)=0_{}$ 

   ⇔ $(x+2)(2x-3)=0_{}$ 

   ⇔ \(\left[ \begin{array}{l}x+2=0\\2x-3=0\end{array} \right.\) \(\left[ \begin{array}{l}x=-2\\x=\frac{3}{2}\end{array} \right.\) 

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2. 1. x(x+1)-x²+1=0

       x²+x-x²+1 = 0

            x+1 = 0

             x= -1

   2. 4x(x-2)-6+3x=0

       4x(x-2)-3(2-x) =0

        4x(x-2) + 3(x-2) = 0

        (x-2)(4x+3) = 0

   -TH1: x-2=0 –>x=2

   – TH2: 4x+3=0 –> x=-3/4

   3. x(x+2)-3(x+2)=0

        (x+2)(x-3)=0

   – TH1: x+2=0 –> x=-2

   – TH2: x-3 = 0 –> x=3

   4. 8x(x-5)-2x+10=0

       8x(x-5)-2(x-5)=0

          (x-5)(8x-2)=0

   -Th1: x-5=0 –> x=5

   -Th2: 8x-2=0 –> x= 1/4

   5..x(2x-3)-2(3-2x)=0

     x(2x-3)+2(2x-3)=0

     (2x-3)(x+2)=0

   -TH1: 2x-3=0–> X=3/2

   -TH2:x+2=0–> x=-2
   

           NHỚ VOTE 5 SAO CHO MIK NHÉ. THANK YOU!!

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