Tìm \(x\) , biết: a) \(\left( 3x-1 \right)\left( 2...

A

- Hướng dẫn giải

Phương pháp giải:

- Chuyển vế đưa về dạng \(a.b=0\). Khi đó \(a=0\) hoặc \(b=0\).

Giải chi tiết:

\(\begin{array}{l}{\rm{a)}}\left( {3x - 1} \right)\left( {2x + 7} \right) - \left( {x + 1} \right)\left( {6x - 5} \right) = 16\\\left( {6{x^2} - 2x + 21x - 7} \right) - \left( {6{x^2} + 6x - 5x - 5} \right) - 16 = 0\\6{x^2} - 2x + 21x - 7 - 6{x^2} - 6x + 5x + 5 - 16 = 0\\18x - 18 = 0\\18x = 18\\x = 1\end{array}\)

\(\begin{array}{l}{\rm{b)}}{\left( {2x + 3} \right)^2} - 2\left( {2x + 3} \right)\left( {2x - 5} \right) + {\left( {2x - 5} \right)^2} = {x^2} + 6x + 64\\{\left[ {\left( {2x + 3} \right) - \left( {2x - 5} \right)} \right]^2} - \left( {{x^2} + 6x + 64} \right) = 0\\{\left( {2x + 3 - 2x + 5} \right)^2} - \left( {{x^2} + 6x + 64} \right) = 0\\64 - {x^2} - 6x - 64 = 0\\ - {x^2} - 6x = 0\\ - x\left( {x + 6} \right) = 0\\\left[ \begin{array}{l}x = 0\\x + 6 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = - 6\end{array} \right.\end{array}\)

\(\begin{array}{l}{\rm{c)}}\left( {{x^4} + 2{x^3} + 10x - 25} \right):\left( {{x^2} + 5} \right) = 3\\{x^2} + 2x - 5 = 3\\{x^2} - 2x - 8 = 0\\{x^2} - 4x + 2x - 8 = 0\\x\left( {x - 4} \right) + 2\left( {x - 4} \right) = 0\\\left( {x - 4} \right)\left( {x + 2} \right) = 0\\\left[ \begin{array}{l}x - 4 = 0\\x + 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 4\\x = - 2\end{array} \right.\end{array}\)

Vậy \(x=4\) hoặc \(x=-2\).

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