4) $(2x+ 7)(x- 5)(5x +1) = 0$
$<=> 2x + 7 =0$
$hoặc$ $x - 5=0$
$hoặc$ $5x+1 = 0$
$<=> x = -7/2$
$hoặc$ $x = 5$
$hoặc$ $x = -1/5$
.
5) $2x(x-3) + 5(x- 3) = 0$
$<=> (x-3)(2x+5) =0$
<=>\(\left[ \begin{array}{l}x-3=0\\2x+5=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=3\\x=-5/2\end{array} \right.\)
.
6) $(x² -4) + (x-2)(3-2x)=0$
$<=> (x-2)(x+2)+ (x-2)(3-2x)=0$
$<=> (x-2)(x+2+3-2x) = 0$
$<=> (x-2)(5-x) = 0$
<=> \(\left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
.
7) $x³-3x²+3x -1 = 0$
$<=> (x-1)³ = 0$
$<=> x-1 = 0$
$<=> x= 1$
.
8) $x(2x - 7)- 4x+14 = 0$
$<=> 2x² - 7x - 4x + 14 = 0$
$<=> (2x² - 4x) - (7x-14) =0$
$<=> 2x(x-2) - 7(x-2) = 0$
$<=> (x-2)(2x-7) = 0$
<=> \(\left[ \begin{array}{l}x-2=0\\2x-7=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=2\\x=7/2\end{array} \right.\)
.
9) $(2x-5)² -(x+2)² = 0$
$<=> (2x-5-x-2)(2x-5+x+2) = 0$
$<=> (x-7)(3x-3) = 0$
<=> \(\left[ \begin{array}{l}x-7=0\\3x-3=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=7\\x=1\end{array} \right.\)
.
10) $x² -x-(3x-3) = 0$
$<=> x² - x - 3x + 3 =0$
$<=> x(x-1) - 3(x-1) = 0$
$<=> (x-1)(x-3) = 0$
<=> \(\left[ \begin{array}{l}x-1=0\\x-3=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=1\\x=3\end{array} \right.\)
.
11) $x(2x -9) = 3x(x-5)$
$<=> 2x² - 9x = 3x² - 15x$
$<=> x² - 6x =0$
$<=> x(x-6) = 0$
<=>\(\left[ \begin{array}{l}x=0\\x-6=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)
.
12)$ 0.5x(x-3) = (x-3)(1.5x-1)$
$<=> 0.5x(x-3) - (x-3)(1.5x-1) = 0$
$<=> (x-3)(0,5x - 1,5x + 1) = 0$
$<=> (x-3)(-x+1)=0$
<=>\(\left[ \begin{array}{l}x-3=0\\-x+1=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)
.
13) $3x-15 = 2x(x-5)$
$<=> 3x-15 - 2x(x-5) = 0$
$<=> 3(x-5) - 2x(x-5) = 0$
$<=> (x-5)(3-2x) = 0$
<=> \(\left[ \begin{array}{l}x-5=0\\3-2x=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=5\\x=3/2\end{array} \right.\)
.
14) $\dfrac{3}{7}$$x$ $-1$ = $\dfrac{1}{7}$ $(3x-7)$
<=> $\dfrac{3}{7}$$x$ $-1$ = $\dfrac{3}{7}$$x$ $-$$1$
$\text{ => PT có vô số nghiệm}$
.
15) $(x² - 2x +1)-4 = 0$
$<=> (x-1)² - 2² = 0$
$<=> (x-1-2)(x-1+2) = 0$
$<=> (x-3)(x+1) = 0$
<=> \(\left[ \begin{array}{l}x-3=0\\x+1=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
.
16) $x² -x = -2x +2$
$<=> x² - x + 2x - 2 =0$
$<=> x(x-1) + 2(x-1) = 0$
$<=> (x-1)(x+2) = 0$
<=> \(\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
.
17) $4x² +4x+1 =x²$
$<=> 3x² + 4x + 1 = 0$
$<=> 3x² + 3x + x + 1 = 0$
$<=> 3x(x+1) + (x+1) = 0$
$<=> (x+1)(3x+1) = 0$
<=> \(\left[ \begin{array}{l}x+1=0\\3x+1=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-1\\x=-1/3\end{array} \right.\)
.
18) $x²- 5x+6 = 0$
$<=> x² -3x - 2x + 6 = 0$
$<=> x(x-3) - 2(x-3) = 0$
$<=> (x-3)(x-2) = 0$
<=>\(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\)
.
19) $2x³+6x² = x²+3x$
$<=> 2x³ + 6x² - x² - 3x = 0$
$<=> 2x²(x+3) - x(x+3) = 0$
$<=> (x+3)(2x²-x) = 0$
<=>\(\left[ \begin{array}{l}x+3=0\\2x²-x=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-3\\x=0 và x=1/2\end{array} \right.\)
.
20)$ (3x-1)(x² +2) = (3x- 1)(7x-10)$
$<=> (3x-1)(x² +2) - (3x- 1)(7x-10) = 0$
$<=> (3x-1)(x² + 2 - 7x + 10) = 0$
$<=> (3x-1)(x² - 7x + 12) =0$
<=>\(\left[ \begin{array}{l}3x-1=0\\x² - 7x + 12=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=1/3\\ x=4 và x=3\end{array} \right.\)
.