Đáp án:
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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``}`
