2 TÌM X a, ( 2x + 4 )^30 =0 , b, x^10 =x , c, x ^20 – x^2 =0 , d, x^15 =x , e, (x-5)^4 = (x-5) ^6

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1. Đáp án:

    

   Giải thích các bước giải:

   `a,(2x+4)^30=0`

   `→2x+4=0`

   `→2x=-4`

   `→x=-2`

   `b,x^10 =x`

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

   `c,x^20 -x^2=0`

   `→x^20=x^2`

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

   `d,x^15 =x`

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

   `e,(x-5)^4 =(x-5)^6`

   `to`\(\left[ \begin{array}{l}x-5=0\\x-5=-1\\x-5=1\end{array} \right.\) 

   `to`\(\left[ \begin{array}{l}x=5\\x=4\\x=6\end{array} \right.\) 

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2. $a$) $(2x+4)^{30} = 0$

   $⇔ 2x+4 = 0$

   $⇔ 2x = -4$

   $⇔ x = -2$

      Vậy $x=-2$.

   $b$) $x^{10} = x$

   $⇔ x^{10} – x =0$

   $⇔ x.(x^{9} – 1) = 0$

   $⇒$ \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\) 

      Vậy $x$ $∈$ `{0;1}`.

   $c$) $x^{20} – x^2 = 0$

   $⇔ x^2.(x^{18} – 1) = 0$

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

      Vậy $x$ $∈$ `{0;±1}`.

   $d$) $x^{15} =x$

   $⇔ x^{15} – x = 0$

   $⇔ x.(x^{14} – 1) = 0$

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

     Vậy $x$ $∈$ `{0;±1}`.

   $e$) $(x-5)^4 = (x-5) ^64$

   $⇔ (x-5)^4 – (x-5)^6 = 0$

   $⇔ (x-5)^4.[1 – (x-5)^2] = 0$

   $⇒$ \(\left[ \begin{array}{l}x-5=0\\x-5=1\\x-5=-1\end{array} \right.\) 

   $⇔$ \(\left[ \begin{array}{l}x=5\\x=6\\x=4\end{array} \right.\) 

       Vậy $x$ $∈$ `{4;5;6}`.

    

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