Tìm x a) $(x-3)^{3}$ + 3 – x = 0 b) x + $6x^{2}$ = 0 c) 5x ( x- 2) – ( 2 – x) = 0 d) ( x+ 1) = $(x+1)^{2}$ ANH CHỊ GIÚP EM VS E ĐNG CẦN GẤP

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1. Đáp án:

   a, Ta có : 

   `(x – 3)^2 + 3 – x = 0`

   ` <=> (x – 3)^2 – (x – 3) = 0`

   ` <=> (x – 3)(x – 3 – 1) = 0`

   ` <=> (x – 3)(x – 4) = 0`

   <=>  \(\left[ \begin{array}{l}x – 3 = 0\\x – 4 = 0\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}x=3\\x=4\end{array} \right.\) 

   b, `x + 6x^2 = 0`

   ` <=> x(1 + 6x) = 0`

   <=> \(\left[ \begin{array}{l}x=0\\1 + 6x = 0\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}x=0\\x=-1/6\end{array} \right.\) 

   c, `5x(x – 2) – (2 – x) = 0`

   ` <=> 5x(x – 2) + (x – 2) = 0`

   ` <=> (x – 2)(5x + 1) = 0`

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

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

   d, Ta có : 

   `(x + 1) = (x + 1)^2`

   ` <=> (x + 1)^2 – (x + 1) = 0`

   ` <=> (x + 1)(x + 1 – 1) = 0`

   ` <=> (x + 1).x = 0`

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

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

   Giải thích các bước giải:

    

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2. Đáp án:

    $a)
    {\left[\begin{aligned}x=3\\x=4\end{aligned}\right.}\\
   b)
   {\left[\begin{aligned}x=0\\x=\frac{-1}{6}\end{aligned}\right.}\\
   c)
    {\left[\begin{aligned}x=2\\x=\frac{-1}{5}\end{aligned}\right.}\\
   d)
    {\left[\begin{aligned}x=-1\\x=0\end{aligned}\right.}$

   Giải thích các bước giải:

    $a)
   (x-3)^2+3-x=0\\
   \Leftrightarrow (x-3)^2-(x-3)=0\\
   \Leftrightarrow (x-3)(x-3-1)=0\\
   \Leftrightarrow (x-3)(x-4)=0\\
   \Leftrightarrow {\left[\begin{aligned}x-3=0\\x-4=0\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x=3\\x=4\end{aligned}\right.}\\
   b)
   x+6x^2=0\\
   \Leftrightarrow x(1+6x)=0\\
   \Leftrightarrow {\left[\begin{aligned}x=0\\1+6x=0\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x=0\\6x=-1\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x=0\\x=\frac{-1}{6}\end{aligned}\right.}\\
   c)
   5x(x-2)-(2-x)=0\\
   \Leftrightarrow 5x(x-2)+(x-2)=0\\
   \Leftrightarrow (x-2)(5x+1)=0\\
   \Leftrightarrow {\left[\begin{aligned}x-2=0\\5x+1=0\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x=2\\5x=-1\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x=2\\x=\frac{-1}{5}\end{aligned}\right.}\\
   d)
   (x+1)=(x+1)^2\\
   \Leftrightarrow (x+1)-(x+1)^2=0\\
   \Leftrightarrow (x+1)(1-x-1)=0\\
   \Leftrightarrow (x+1).(-x)=0\\
   \Leftrightarrow {\left[\begin{aligned}x+1=0\\-x=0\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x=-1\\x=0\end{aligned}\right.}$

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