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$B$
$B1$
$a,3x+1=7x-11$
$ĐKXĐ:x∈R$
pt$⇔7x-3x=1+11$
$⇔4x=12$
$⇔x=3$
Vậy...
$b,2(x+1)=3+2x$
$⇔2x+2-2x-3=0$
$⇔-1=0$(vô $n_o$)
$c,5-(6-x)=4(3-2x)$
$⇔5-6+x=12-8x$
$⇔x+8x=12+6-5$
$⇔9x=13$
$⇔x=\dfrac{13}{9}$
$d,2(x-3)+5x(x-1)=5x^2$
$⇔2x-6+5x^2-5x-5x^2=0$
$⇔-3x-6=0$
$⇔-3x=6$
$⇔x=-2$
$B2$
$a,\dfrac{2.(x+3)}{3}=\dfrac{5-4x}{2}$
$⇔4(x+3)=3(5-4x)$
$⇔4x+12-15+12x=0$
$⇔16x=3$
$⇔x=\dfrac{3}{16}$
$b,\dfrac{5x+3}{12}=\dfrac{1+2x}{9}$
$⇔4(5x+3)=3(1+2x)$
$⇔20x+12-3-6x=0$
$⇔14x=-9$
$⇔x=\dfrac{-9}{14}$
$c,\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}$
$⇔3(x-3)=5(18-1+2x)$
$⇔3x-9=85+10x$
$⇔7x=-94$
$⇔x=\dfrac{-94}{7}$
$d,x-\dfrac{x+1}{3}=\dfrac{2x+1}{5}$
$⇔5(3x-x-1)=3(2x+1)$
$⇔10x-5=6x+3$
$⇔4x=8$
$⇔x=2$
$B3$
$a,(x+2)(x-3)=0$
$⇔$\(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
$b,(2x+3)(-x+7)=0$
$⇔$\(\left[ \begin{array}{l}2x+3=0\\-x+7=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{-3}{2}\\x=7\end{array} \right.\)
$c,(x+6)(3x-1)+x+6=0$
$⇔(x+6)(3x-1+1)=0$
$⇔$\(\left[ \begin{array}{l}x+6=0\\3x=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-6\\x=0\end{array} \right.\)
$d,(x-2)(x+1)=x^2-4$
$⇔(x-2)(x+1)=(x-2)(x+2)$
$⇔(x-2)(x+1)-(x-2)(x+2)=0$
$⇔(x-2)(x+1-x-2)=0$
$⇔x-2=0$
$⇔x=2$
