Giải câu 4 trang 57 toán VNEN 9 tập 2.
a) $x^3-5x^2-2x+10=0$
$\Leftrightarrow x(x^2-2)-5(x^2-2)=0$
$\Leftrightarrow (x^2-2)(x-5)=0$
$\Leftrightarrow \left[ \begin{matrix}x^2=2\\ x = 5\end{matrix}\right.$
$\Leftrightarrow \left[ \begin{matrix}x = \pm \sqrt{2}\\ x = 5\end{matrix}\right.$
b) $x^5+2x^3-x^2-2=0$
$\Leftrightarrow x^3(x^2+2)-(x^2+2) = 0$
$\Leftrightarrow (x^2+2)(x^3-1) = 0$
$\Leftrightarrow \left[ \begin{matrix}x^2+2=0\\ x^3-1=0\end{matrix}\right.$
$\Leftrightarrow \left[ \begin{matrix}x^2=-2(vô lí) \\ x=1\end{matrix}\right.$
c) $(2x^2-5x+1)^2 = (x^2-5x+6)^2$
$\Leftrightarrow (2x^2-5x+1)^2 - (x^2-5x+6)^2=0$
$\Leftrightarrow (x^2-5)(3x^2-10x+7)=0$
$\Leftrightarrow \left[ \begin{matrix}x^2-5=0\;\; (1)\\ 3x^2-10x+7=0\;\;(2)\end{matrix}\right.$
Giải (1):
$x^2-5 = 0 \Leftrightarrow x =\pm \sqrt{5}$
Giải (2):
$\Delta' = (-5)^2-3\times 7=4 \Rightarrow \sqrt{\Delta} = 2$
$\Rightarrow \left[ \begin{matrix}x_1=1\\ x_2=\frac{7}{3} \end{matrix}\right.$
d) $(2x^2-3)^2-4(x-1)^2=0$
$\Leftrightarrow (2x^2-3)^2-(2x-2)^2=0$
$\Leftrightarrow (2x^2-2x-1)(2x^2+2x-5)=0$
$\Leftrightarrow \left[ \begin{matrix}2x^2-2x-1=0\;\; (3)\\ 2x^2+2x-5=0\;\;(4)\end{matrix}\right.$
Giải (3)
$\Delta' = (-1)^2-2\times (-1) = 3 \Rightarrow \sqrt{\Delta }=\sqrt{3}$
$\Rightarrow \left[ \begin{matrix}x_1 = \frac{1+\sqrt{3}}{2}\\ x_2 = \frac{1-\sqrt{3}}{2}\end{matrix}\right.$
Giải (4):
$\Delta' = 1^2-2\times (-5) = 11 \Rightarrow \sqrt{\Delta }=\sqrt{3}$
$\Rightarrow \left[ \begin{matrix}x_1 = \frac{-1+\sqrt{11}}{2}\\ x_2 = \frac{-1-\sqrt{11}}{2}\end{matrix}\right.$