a)3x+9=0 b)(x-1)^2-4=0 c)x+3 phần x+1+x-2phan x=1. d)|x+2|=10 e)2(1-2x)>=4-5x

0 bình luận về “a)3x+9=0 b)(x-1)^2-4=0 c)x+3 phần x+1+x-2phan x=1. d)|x+2|=10 e)2(1-2x)>=4-5x”

1. `a, 3x + 9 = 0`

   `⇔ 3(x + 3) = 0`

   `⇔ x + 3 = 0`

   `⇔ x = -3`

   `b, (x – 1)^2 – 4 = 0`

   `⇔ (x – 1)^2 – 2^2 = 0`

   `⇔ (x – 1 – 2)(x – 1 + 2) = 0`

   `⇔ (x – 3)(x + 1) = 0`

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

   `c, \frac{x+3}{x+1} + \frac{x-2}{x} = 1` `(đk: x \ne 0; x \ne -1)`

   `⇔ \frac{x(x+3)}{x(x+1)} + \frac{(x-2)(x+1)}{x(x+1)} = \frac{x+1}{x+1}`

   `⇒ x(x + 3) + (x – 2)(x + 1) = x + 1`

   `⇔ x^2 + 3x + x^2 + x – 2x – 2 = x + 1`

   `⇔ 2x^2 + x – 3 = 0`

   `⇔ 2x^2 + 3x – 2x – 3 = 0`

   `⇔ 2x(x – 1) + 3(x – 1) = 0`

   `⇔ (2x + 3)(x – 1) = 0`

   `⇔` \(\left[ \begin{array}{l}2x+3=0\\x-1=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{-3}{2}(TM)\\x=1(TM)\end{array} \right.\) 

   `d, |x + 2| = 10`

   `⇔` \(\left[ \begin{array}{l}x+2=10\\x+2=-10\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=8\\x=-12\end{array} \right.\) 

   `e, 2(1 – 2x) ≥ 4 – 5x`

   `⇔ 2 – 4x ≥ 4 – 5x`

   `⇔ x ≥ 2`

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2. a) 3x+9=0

   <=> 3x=-9

   <=>x=-3

   Vậy PT có nghiệm là x=-3

   b) (x-1)²-4=0

   <=>x²-2x+1-4=0

   <=>x²-2x-3=0

   <=>x²-x+3x-3=0

   <=>x(x-1)+3(x-1)=0

   <=>(x-1)(x+3)=0

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

   Vậy PT có nghiệm x=1 và x=-3

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