Giải câu 6.5 trang 68 toán VNEN 9 tập 2

a) $(4x^2-25)(2x^2-7x-9)=0$

$\Leftrightarrow \left[ \begin{matrix}4x^2-25=0\\ 2x^2-7x-9=0\end{matrix}\right.$

$\Leftrightarrow \left[ \begin{matrix}x^2=\frac{25}{4}\\ \Delta = (-7)^2 -4\times 2\times (-9) = 121 \Rightarrow \sqrt{\Delta} = 11\end{matrix}\right.$

$\Leftrightarrow \left[ \begin{matrix}x=\pm \frac{5}{2}\\ x = \frac{9}{2}\\ x = -1\end{matrix}\right.$

b) $(2x^2-3)^2-4(x-1)^2=0$

$\Leftrightarrow [(2x^2-3) + 2(x-1)][(2x^2-3)-(2x-2)] = 0$

$\Leftrightarrow (2x^2+2x-5)(2x^2-2x-1) = 0$

$\Leftrightarrow \left[ \begin{matrix}2x^2+2x-5=0\\ 2x^2-2x-1=0\end{matrix}\right.$

$\Leftrightarrow \left[ \begin{matrix}\Delta' = 1^2 - 2\times (-5) = 11 \Rightarrow \sqrt{\Delta} = \sqrt{11}\\ \Delta' = (-1)^2 - 2\times (-1) = 3 \sqrt{\Delta} = \sqrt{3}\end{matrix}\right.$

$\Leftrightarrow \left[ \begin{matrix}x_{1,2}=\frac{-1\pm \sqrt{11}}{2}\\ x_{3,4} =\frac{1\pm \sqrt{3}}{2}\end{matrix}\right.$

c) $x^3+3x^2+x+3=0$

$\Leftrightarrow (x^2+1)(x+3) =0$

$\Leftrightarrow \left[ \begin{matrix}x^2+1 = 0\\ x+3 =0 \end{matrix}\right.$

$\Leftrightarrow x = -3$

d) $x^3+8-4x^2-2x=0$

$\Leftrightarrow x^2(x-4)-2(x-4) = 0$

$\Leftrightarrow (x^2-2)(x-4) = 0$

$\Leftrightarrow \left[ \begin{matrix}x^2-2=0\\ x-4=0\end{matrix}\right.$

$\Leftrightarrow \left[ \begin{matrix}x=\pm \sqrt{2}\\ x=4\end{matrix}\right.$