Tìm x: (x+2)^2-(x+2)(x-2)=0 x^2-7x+10=0 x^3+3x=3x^2+1 7x^2-28=0 2/3x(x^2-4)=0 2x.(3x-5)-(5-3x)=0

2 bình luận về “Tìm x: (x+2)^2-(x+2)(x-2)=0 x^2-7x+10=0 x^3+3x=3x^2+1 7x^2-28=0 2/3x(x^2-4)=0 2x.(3x-5)-(5-3x)=0”

1. Giải đáp: $\begin{array}{l}
   a)x =  – 2\\
   b)x = 2;x = 5\\
   c)x = 1\\
   d)x = 2;x =  – 2\\
   e)x = 0;x =  – 2;x = 2\\
   f)x = \dfrac{5}{3};x =  – \dfrac{1}{2}
   \end{array}$

    

   Lời giải và giải thích chi tiết:

   $\begin{array}{l}
   a){\left( {x + 2} \right)^2} – \left( {x + 2} \right)\left( {x – 2} \right) = 0\\
    \Leftrightarrow \left( {x + 2} \right)\left( {x + 2 – x + 2} \right) = 0\\
    \Leftrightarrow \left( {x + 2} \right).4 = 0\\
    \Leftrightarrow x + 2 = 0\\
    \Leftrightarrow x =  – 2\\
   Vậy\,x =  – 2\\
   b){x^2} – 7x + 10 = 0\\
    \Leftrightarrow {x^2} – 2x – 5x + 10 = 0\\
    \Leftrightarrow \left( {x – 2} \right)\left( {x – 5} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   x – 2 = 0\\
   x – 5 = 0
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = 2\\
   x = 5
   \end{array} \right.\\
   Vậy\,x = 2;x = 5\\
   c){x^3} + 3x = 3{x^2} + 1\\
    \Leftrightarrow {x^3} – 3{x^2} + 3x – 1 = 0\\
    \Leftrightarrow {\left( {x – 1} \right)^3} = 0\\
    \Leftrightarrow x – 1 = 0\\
    \Leftrightarrow x = 1\\
   Vậy\,x = 1\\
   d)7{x^2} – 28 = 0\\
    \Leftrightarrow 7{x^2} = 28\\
    \Leftrightarrow {x^2} = 4\\
    \Leftrightarrow x = 2;x =  – 2\\
   Vậy\,x = 2;x =  – 2\\
   e)\dfrac{2}{3}x\left( {{x^2} – 4} \right) = 0\\
    \Leftrightarrow \dfrac{2}{3}x\left( {x – 2} \right)\left( {x + 2} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = 0\\
   x – 2 = 0\\
   x + 2 = 0
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = 0\\
   x = 2\\
   x =  – 2
   \end{array} \right.\\
   Vậy\,x = 0;x =  – 2;x = 2\\
   f)2x\left( {3x – 5} \right) – \left( {5 – 3x} \right) = 0\\
    \Leftrightarrow 2x\left( {3x – 5} \right) + \left( {3x – 5} \right) = 0\\
    \Leftrightarrow \left( {3x – 5} \right)\left( {2x + 1} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   3x – 5 = 0\\
   2x + 1 = 0
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = \dfrac{5}{3}\\
   x = \dfrac{{ – 1}}{2}
   \end{array} \right.\\
   Vậy\,x = \dfrac{5}{3};x =  – \dfrac{1}{2}
   \end{array}$

   Trả lời
2. 1.(x+2)^2-(x+2)(x-2)=0

   =>(x+2)[(x+2)-(x-2)]=0

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

   =>4(x+2)=0

   =>x+2=0

   =>x=-2

   2.x^2-7x+10=0

   =>x^2-2x-5x+10=0

   =>x(x-2)-5(x-2)=0

   =>(x-5)(x-2)=0

   => \(\left[ \begin{array}{l}x-5=0\\x-2=0\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=5\\x=2\end{array} \right.\) 

   3.x^3+3x=3x^2+1

   =>x^3-3x^2+3x-1=0

   =>(x-1)^3=0

   =>x-1=0

   =>x=1

   4.7x^2-28=0

   =>7(x^2-4)=0

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

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

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

   5. 2/3 x(x^2-4)=0

   =>2/3 x(x-2)(x+2)=0

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

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

   6.2x(3x-5)-(5-3x)=0

   =>2x(3x-5)+(3x-5)=0

   =>(2x+1)(3x-5)=0

   => \(\left[ \begin{array}{l}2x+1=0\\3x-5=0\end{array} \right.\) 

   => \(\left[ \begin{array}{l}2x=-1\\3x=5\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=\dfrac{5}{3}\end{array} \right.\) 

   Trả lời