Giải pt tích sau đây : a, (x+1)(2x-3)=0 b, (5x-1)(3-2x)(x-1)=0 c, x^2-1+(x+1)(2x-4)=0

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1. `a, (x+1)(2x-3)=0`

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

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

   Vậy `S = {-1,3/2}`

   `b, (5x-1)(3-2x)(x-1)=0`

   `<=> 5x-1=0` hoặc `3-2x=0` hoặc `x-1=0`

   `<=> x=1/5` hoặc `x=3/2` hoặc `x=1`

   Vậy `S = {1/5,3/2,1}`

   `c, x^2-1+(x+1)(2x-4)=0`

   `<=>(x-1)(x+1)+(x+1)(2x-4)=0`

   `<=>(x+1)(x-1+2x-4)=0`

   `<=>(x+1)(3x-5)=0`

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

   `<=>`\(\left[ \begin{array}{l}x=-1\\x=5/3\end{array} \right.\)

   Vậy `S = {-1,5/3}`

    

   Trả lời

2. Đáp án:

   ↓↓↓↓

   Giải thích các bước giải:

    a) `(x+1)(2x-3) = 0`

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

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

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

   Vậy nghiệm pt `S ={-1;“3/2“}`

   b) `(5x-1)(3-2x)(x-1) = 0`

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

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

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

   Vậy `S ={“1/5`;`3/2`;`1“}`

   c) `x^2-1+(x+1)(2x-4) = 0`

   ⇔ `(x+1)(x-1)+(x+1)(2x-4) = 0`

   ⇔ `(x+1)(x-1+2x-4) = 0`

   ⇔ `(x+1)(3x-5) = 0`

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

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

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

   Vậy `S ={-1;“5/3“}`

   Trả lời