Đáp án:
a) 6x² - (2x+5)(3x-2) = 0
⇔6x²-6x²+4x-15x+10=0
⇔-11x=-10
⇔x=$\frac{10}{11}$
b, 5x(x-2004) - 2x+4008 = 0
⇔5x²-10020x-2x+4008=0
⇔5x²-10022x+4008=0
⇔\(\left[ \begin{array}{l}x=2004\\x=\frac{2}{5}\end{array} \right.\)
c,(3x-1)² - 25 = 0
⇔(3x-1-5).(3x-1+5)=0
⇔(3x-6).(3x+4)=0
⇔\(\left[ \begin{array}{l}3x-6=0\\3x+4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=\frac{-4}{3}\end{array} \right.\)
d,(2x²-7x) - 4x+14 =0
⇔2x²-11x+14=0
⇔\(\left[ \begin{array}{l}x=2\\x=\frac{7}{2}\end{array} \right.\)
e,$\frac{2x-3}{4}-\frac{4x-5}{3}=\frac{5-x}{6}$
⇔$\frac{3.(2x-3)}{12}-\frac{4.(4x-5)}{12}=\frac{2.(5-x)}{12}$
⇔$\frac{3.(2x-3)-4.(4x-5)}{12}= \frac{10-2x}{12}$
⇔$\frac{6x-9-16x+20}{12}= \frac{10-2x}{12}$
⇔$\frac{11-10x}{12}= \frac{10-2x}{12}$
⇔$11-10x=10-2x$
⇔$11-10=-2x+10x$
⇔$1=8x$
⇔$x=\frac{1}{8}$
