Giải bài 55 trang 25 – SGK Toán lớp 8 tập 1
Tìm \(x,\) biết:
a) \(x^3 - \dfrac{1}{4}x = 0\)
b) \((2x - 1)^2 - (x + 3)^2 = 0\)
c) \(x^2 (x - 3) + 12 - 4x = 0\)
a) \(x^3 - \dfrac{1}{4}x = 0\)
\(\Leftrightarrow x\left(x^2 - \dfrac{1}{4}\right) = 0\)
\(\Leftrightarrow x\left(x - \dfrac{1}{2}\right)\left(x + \dfrac{1}{2}\right) = 0 \)
\(\Leftrightarrow \left[\begin{array}{l}x = 0 \\ x - \dfrac{1}{2} = 0 \\ x + \dfrac{1}{2} = 0\end{array} \right.\\ \\ \Leftrightarrow \left[\begin{array}{l}x = 0 \\ x = \dfrac{1}{2} \\ x =- \dfrac{1}{2} \end{array} \right.\)
b) \((2x - 1)^2 - (x + 3)^2 = 0\)
\(\Leftrightarrow [(2x - 1) + (x + 3)][(2x - 1) - (x + 3)] = 0\)
\(\Leftrightarrow (2x - 1 + x+ 3)(2x - 1 - x - 3) = 0\)
\(\Leftrightarrow (3x + 2)(x - 4) = 0\)
\(\Rightarrow \left[\begin{array}{l}3x + 2 = 0 \\ x - 4 = 0\end{array} \right. \\ \\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{2}{3} \\ x = 4\end{array} \right.\)
c) \(x^2 (x - 3) + 12 - 4x = 0\)
\(\Leftrightarrow x^2(x - 3) - 4( x - 3) = 0\)
\(\Leftrightarrow (x - 3)(x^2 - 4) = 0\)
\(\Leftrightarrow (x - 3)(x - 2)(x + 2) = 0\)
\(\Rightarrow \left[\begin{array}{l}x - 3 = 0 \\ x + 2 = 0 \\ x - 2 = 0\end{array} \right. \\ \\ \Leftrightarrow \left[\begin{array}{l}x = 3 \\ x = -2 \\ x =2 \end{array} \right.\)
Lưu ý: \( A.B.C = 0 \Leftrightarrow \left[\begin{array}{l} A = 0 \\ B = 0 \\ C = 0\end{array} \right.\)