a (2x-3).(6-2x)=0 b (x+1 phần 2 ) .(2-3x)=0 c 2x(x^2+1).(x^2-9)=0

0 bình luận về “a (2x-3).(6-2x)=0 b (x+1 phần 2 ) .(2-3x)=0 c 2x(x^2+1).(x^2-9)=0”

1. $\text {a, (2x-3)(6-2x) =0}$

   $\text {⇔\(\left[ \begin{array}{l}2x-3 =0\\6-2x =0\end{array} \right.\) }$

   $\text {⇔\(\left[ \begin{array}{l}2x=3\\2x=6\end{array} \right.\) }$

   $\text {⇔\(\left[ \begin{array}{l}x=\frac{3}{2} \\x=3\end{array} \right.\) }$

   $\text {Vậy x=$\frac{3}{2}$ hoặc x=3}$ 

   $\text {b, (x+$\frac{1}{2}$)(2-3x) =0 }$

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

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

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

   $\text {Vậy x=$\frac{-1}{2}$ hoặc x=$\frac{2}{3}$ }$

   $\text {c, 2x(x²+1)(x²-9) =0}$

   $\text {⇔\(\left[ \begin{array}{l}2x =0\\x²+1 =0\\x²-9 =0\end{array} \right.\) }$

   $\text {⇔\(\left[ \begin{array}{l}x =0\\x²+1 =0 (vô nghiệm)\\x²=9\end{array} \right.\) }$

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

   $\text {Vậy x=0 hoặc x=3 hoặc x=-3}$

   $\text {Chúc bạn học tốt~}$

   $\text {@lamtung2}$

     

   Trả lời

2. Giải thích các bước giải:

   `a)`
   `(2x-3).(6-2x)=0`
   TH`1`
   `2x-3=0`
   `=>2x=0+3`
   `=>2x=3`
   `=>x=3:2`
   `=>x=3/2`
   TH`2`
   `6-2x=0`
   `=>2x=6-0`
   `=>2x=6`
   `=>x=6:2`
   `=>x=3`
   Vậy `x\in{3/2;3}`
   `b (x+1 /2 ) .(2-3x)=0`
   TH`1`
   `x+1/2=0`
   `=>x=0-1/2`
   `=>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}`
   `c)`
   `2x(x^2+1).(x^2-9)=0`
   TH`1`
   `2x=0`
   `=>x=0`
   TH`2`
   `x^2+1=0`
   `=>x^2=0-1`
   `=>x^2=-1`
   Vì `x^2ge0` mọi `x`
   `=>x` thuộc rỗng
   TH`3`
   `x^2-9=0`
   `=>x^2=0+9`
   `=>x^2=9`
   `=>x^2=(+-3)^2`
   `=>x=+-3`
   Vậy `x\in{0;+-3}`

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