Giải pt sau (x+1)(x+3)(x+5)(x-6)=0 (2x+1)(3x-2)(5x-8)(2x-1)=0

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1. $(x+1)(x+3)(x+5)(x-6)=0$

   $⇔$\(\left[ \begin{array}{l}x+1=0\\x+3=0\\x+5=0\\x-6=0\end{array} \right.\) 

   $⇔$\(\left[ \begin{array}{l}x=-1\\x=-3\\x-5\\x=6\end{array} \right.\) 

   Vậy: `S={-1;-3;-5;6}`

   $(2x+1)(3x-2)(5x-8)(2x-1)=0$

   $⇔$\(\left[ \begin{array}{l}2x+1=0\\3x-2=0\\5x-8=0\\2x-1=0\end{array} \right.\) 

   $⇔$\(\left[ \begin{array}{l}2x=1\\3x=2\\5x=8\\2x=1\end{array} \right.\) 

   $⇔$\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\\x=\dfrac{8}{5}\\x=\dfrac{1}{2}\end{array} \right.\) 

   Vậy: `S={\frac{1}{2};\frac{2}{3};\frac{8}{5};\frac{1}{2}}`

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2. a)  ${(x+1)(x+3)(x+5)(x-6)=0}$

   ⇔ \(\left[ \begin{array}{l}x+1=0\\x+3=0\\x+5=0\\x-6=0\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x=-1\\x=-3\\x=-5\\x=6\end{array} \right.\)

   $\text{Vậy phương trình đã cho có tập nghiệm S=  {-1;-3;-5;6}}$

   b) ${(2x+1)(3x-2)(5x-8)(2x-1)=0}$

   ⇔ \(\left[ \begin{array}{l}2x+1=0\\3x-2=0\\5x-8=0\\2x-1=0\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}2x=-1\\3x=2\\5x=8\\2x=1\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\\x=\dfrac{8}{5}\\x=\dfrac{1}{2}\end{array} \right.\)

   $\text{Vậy phương trình đã cho có tập nghiệm S= }$ {${\dfrac{-1}{2};\dfrac{2}{3};\dfrac{8}{5};\dfrac{1}{2}}$}

    

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