1) (x-3)(5x+12)=0 2) (2x+1)(2-3x)=0 3) 2x^2+5x=0 4) (x-2019)(x+2020)=0 Giúp ạ cần gấp

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1. $\text{Đáp án + Giải thích các bước giải:}$

   `1//(x-3)(5x+12)=0`

   `⇔` \(\left[ \begin{array}{l}x-3=0\\5x+12=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=3\\5x=-12\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=3\\x=-\dfrac{12}{5}\end{array} \right.\) 

   `\text{Vậy}` `S={3;-(12)/(5)}`

   `2//(2x+1)(2-3x)=0`

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

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

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

   `\text{Vậy}` `S={-(1)/(2);(2)/(3)}`

   `3//2x^{2}+5x=0`

   `<=>x(2x+5)=0`

   `⇔` \(\left[ \begin{array}{l}x=0\\2x+5=0\end{array} \right.\) 

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

   `\text{Vậy}` `S={0;-(5)/(2)}`

   `4//(x-2019)(x+2020)=0`

   `<=>` \(\left[ \begin{array}{l}x-2019=0\\x+2020=0\end{array} \right.\) 

   `<=>` \(\left[ \begin{array}{l}x=2019\\x=-2020\end{array} \right.\) 

   `\text{Vậy}` `S={2019;-2020}`

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2. Giải thích các bước giải:

   `1) (x-3)(5x+12)=0`
   TH`1`
   `x-3=0`
   `=>x=0+3`
   `=>x=3`
   TH`2`
   `5x+12=0`
   `=>5x=0-12`
   `=>5x=-12`
   `=>x=-12:5`
   `=>x=-12/5`
   Vậy `x\in{3;-12/5}`
   `2) (2x+1)(2-3x)=0`
   TH`1`
   `2x+1=0`
   `=>2x=0-1`
   `=>2x=-1`
   `=>x=-1:2`
   `=>x=-1/2`
   TH`2`
   `2-3x=0`
   `=>3x=2-0`
   `=>3x=2`
   `=>x=2:3`
   `=>x=2/3`
   Vậy `x\in{-1/2;2/3}`
   `3) 2x^2+5x=0`
   `=>x(2x+5)=0`
   TH`1`
   `x=0`
   TH`2`
   `2x+5=0`
   `=>2x=0-5`
   `=>2x=-5`
   `=>x=-5:2`
   `=>x=-5/2`
   Vậy `x\in{0;-5/2}`
   `4) (x-2019)(x+2020)=0`
   TH`1`
   `x-2019=0`
   `=>x=0+2019`
   `=>x=2019`
   TH`2`
   `x+2020=0`
   `=>x=0-2020`
   `=>x=-2020`
   Vậy `x\in{2019;-2020}`

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