Hướng dẫn giải
a) (2x – 3)2– 49 = 0
⇔ (2x – 3)2– 72= 0
⇔ (2x – 3 – 7)(2x – 3 + 7) = 0
⇔ (2x – 10)(2x + 4) = 0
\( \Leftrightarrow \left[ \begin{array}{l}2x - 10 = 0\\2x + 4 = 0\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = - 2\end{array} \right.\)
Vậy x = 5, x = - 2.
b) 2x(x – 5) – 7(5 – x) = 0
⇔ 2x(x – 5) + 7(x – 5) = 0
⇔ (x – 5)(2x + 7) = 0
\( \Leftrightarrow \left[ \begin{array}{l}x - 5 = 0\\2x + 7 = 0\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = - \frac{7}{2}\end{array} \right.\)
Vậy \(x = - \frac{7}{2}\), x = 5.
c) x2– 3x – 10 = 0
⇔ x2– 5x + 2x – 10 = 0
⇔ x(x – 5) + 2(x – 5) = 0
⇔ (x – 5)(x + 2) = 0
\( \Leftrightarrow \left[ \begin{array}{l}x - 5 = 0\\x + 2 = 0\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = - 2\end{array} \right.\)
Vậy x = 5, x = – 2.
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