$a) A = -4x -5 =0$
$=> -4x = 5$
$ => x = \dfrac{-5}{4}$
$b) B = 3(2x-1) - 2(x+1) = 0$
$=> 6x - 3 - 2x - 2 = 0$
$=> 4x - 5 = 0$
$=> 4x = 5$
$=> x = \dfrac{5}{4}$
$c) (2x^2 -8)(-x^2+1) = 0$
\(\left[ \begin{array}{l}2x^2-8=0\\-x^2+1=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}2x^2=8\\-x^2=-1\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=2\\x=±1\end{array} \right.\)
$d) 3x - x^3 = 0$
$=> x(3 - x^2) = 0$
=>\(\left[ \begin{array}{l}x=0\\3-x^2=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=0\\x^2=3\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=0\\x=±3\end{array} \right.\)
$e) 2x^3 + 4x = 2x( x^2 + 4)$
Ta có $x^2 + 4 > 0$ nên $2x = 0$
$=> x = 0$
$f) x^3 - x^2 + x - 1 = x^2(x-1) + (x-1) = (x^2 + 1)(x-1) $
Ta có$x^2 + 1 > 0$ nên $x - 1 = 0 => x = 1$
g) l$\dfrac{1}{2}x - 3$l $- \dfrac{1}{2} = 0$
Với x $\geq$ 6 có
$\dfrac{1}{2}x - 3 - \dfrac{1}{2} = 0$
$=> \dfrac{1}{2}x = \dfrac{7}{2}$
$=> x = 7$ ( thỏa mãn )
Với $x < 6$ ta có
$3 - \dfrac{1}{2}x - \dfrac{1}{2} = 0$
$ => \dfrac{1}{2}x = \dfrac{5}{2 }$
$ => x = 5 ( thỏa mãn )
h) l$3x -2$l + l$4-6x$l $= 0$
=> $\left \{ {{3x-2=0} \atop {4-6x=0}} \right.$
=> $\left \{ {{x=\dfrac{2}{3}} \atop {x=\dfrac{2}{3}}} \right.$
Vậy $x = \dfrac{2}{3}$
i) l$x-1$l $+ (x^2 -1)^2 = 0$
$=> x - 1 = 0 => x = 1$
Và $x^2 - 1 = 0 => x = ± 1$
Vậy $x= 1$
$j) 4x^2 -3 x +7= 0$
$=> 16x^2 - 12x + 28 = 0$
$=> ( 16x^2 - 12x + 9/4) + 103/4 = 0$
$=> ( 4x - 3/2)^2 + 103/4 = 0$
Mà $( 4x - 3/2)^2 + 103/4 > 0$ => đa thức ko có nghiệm
$k) 7x^2 - 2x - 9 = 0$
$=> x(7x+9) - (7x+9) = 0$
$=> (x-1)(7x+9) = 0 $
$=> x - 1 = 0$ hoặc $7x + 9 = 0$
$=> x= 1$ hoặc $x = -9/7$
$l) 5x^2 - 11x + 6 = 0$
$=> 5x(x-1) - 6(x-1) = 0$
$=> (5x-6)(x-1) = 0$
$ => x = 6/5$ hoặc $x = 1$
