a.(x+4)^2+(3+x)(3-x)+1 b.(x+4)/2-(x-1)/3=1 c.(x-2)^2=9

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1. a)(x+4)²+(3+x)(3-x)=1

   ⇔x²+8x+16+9-x²=1

   ⇔(x²-x²)+8x+16+9=1

   ⇔8x=-24

   ⇔x=-3

   vậy S={-3}

   b)$\frac{x+4}{2}$ -$\frac{x-1}{3}$ =1

   ⇔ $\frac{3(x+4)-2(x-1)}{6}$ =$\frac{6}{6}$ 

   ⇔3x+12-2x+2=6

   ⇔3x-2x=6-12-2

   ⇔x=-8

   vậy S={-8}

   c)(x-2)²=9

   ⇒(x-2)²=3²

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

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

   vậy s={5,-1}

   xin 5sao và tlhn nha bạn:3

   !học tốt nha!

   @Dinosieucute

    

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2. `a)(x+4)^2+(3+x)(3-x)=1`

   `⇔x^2+8x+16+9-x^2=1`

   `⇔(x^2-x^2)+8x+16+9=1`

   `⇔8x=-24`

   `⇔x=-3`

   Vậy `S={-3}`

   `b)(x+4)/2-(x-1)/3=1`

   `⇔(3(x+4)-2(x-1))/6=6/6`

   `⇔3x+12-2x+2=6`

   `⇔3x-2x=6-12-2`

   `⇔x=-8`

   Vậy `S={-8}`

   `c)(x-2)^2=9`

   `→(x-2)^2=3^2`

   `→` \(\left[ \begin{array}{l}x-2=3\\x-2=-3\end{array} \right.\) 

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

   Vậy `S={5;-1}`

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