Câu 55 : Trang 25 - toán 8 tập 1 phần đại số
Tìm x, biết:
a) x3 – \(\frac{1}{4}\)x = 0;
b) (2x – 1)2 – (x + 3)2 = 0;
c) x2(x – 3) + 12 – 4x = 0.
Bài Làm:
a) x3 – \(\frac{1}{4}\)x = 0
<=> x(x2 –\(\left ( \frac{1}{2} \right )^{2}\)) = 0
<=>x(x - \(\frac{1}{2}\))(x + \(\frac{1}{2}\)) = 0
$\Leftrightarrow \left[ \begin{array}{ll} x = 0 \\ x + \frac{1}{2}=0 \\x - \frac{1}{2}= 0\\\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{ll} x = 0 \\ x =-\frac{1}{2} \\x = \frac{1}{2}\\\end{array} \right.$
Vậy x = 0; x = -\(\frac{1}{2}\); x = \(\frac{1}{2}\).
b) (2x – 1)2 – (x + 3)2 = 0
<=>[(2x - 1) - (x + 3)][(2x - 1) + (x + 3)] = 0
<=>(2x - 1 - x - 3)(2x - 1 + x + 3) = 0
<=>(x - 4)(3x + 2) = 0
$\Leftrightarrow \left[ \begin{array}{ll} x - 4 = 0 \\ 3x + 2 =0 \\\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{ll} x = 4 \\ x =-\frac{2}{3} \\\end{array} \right.$
Vậy x = 4; x = -\(\frac{2}{3}\).
c) x2(x – 3) + 12 – 4x = 0
<=> x2(x – 3) - 4(x -3)= 0
<=> (x - 3)(x2- 22) = 0
<=> (x - 3)(x - 2)(x + 2) = 0
$\Leftrightarrow \left[ \begin{array}{ll} x - 3 = 0 \\ x + 2 =0 \\x - 2 = 0 \\\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{ll} x = 3 \\ x =-2 \\x = 2 \\\end{array} \right.$
Vậy x = 3; x = 2; x = -2.