Tìm \(x\) biết:\(a)\ {{\left( x-1 \right)}^{2}}=x-...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Hướng dẫn chi tiết:

\(\begin{array}{l}a){\left( {x - 1} \right)^2} = x - 1\\ \Leftrightarrow \left( {x - 1} \right)\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x - 1 - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 1 = 0\\x - 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = 2\end{array} \right.\end{array}\)

Vậy \(x=1\) hoặc \(x=2\)

\(\begin{array}{l}b)\;7{x^2} + 2x = 0\\ \Leftrightarrow 7x.x + 2.x = 0\\ \Leftrightarrow x\left( {7x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = - \frac{2}{7}\end{array} \right.\end{array}\)

Vậy \(x=0\) hoặc \(x = \frac{{ - 2}}{7}\)

\(\begin{array}{l}c)7{x^2}\left( {x - 7} \right) + 5x\left( {7 - x} \right) = 0\\ \Leftrightarrow 7x.x\left( {x - 7} \right) - 5.x\left( {x - 7} \right) = 0\\ \Leftrightarrow \left( {7x.x - 5.x} \right)\left( {x - 7} \right) = 0\\ \Leftrightarrow x\left( {7x - 5} \right)\left( {x - 7} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\7x - 5 = 0\\x - 7 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = \frac{5}{7}\\x = 7\end{array} \right.\end{array}\)

Vậy \(x=0\) hoặc \(x=7\) hoặc \(x=\frac{5}{7}\)

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

Vậy \(x=1\) hoặc \(x=-1\) hoặc \(x=3\).