Giải thích các bước giải:
Bài 5:
$a)M=(\dfrac{1}{x^2-x}+\dfrac{1}{x-1}):\dfrac{x+1}{x^2+1-2x}$
$=(\dfrac{1}{x(x-1)}+\dfrac{x}{x(x-1)}).\dfrac{(x-1)^2}{x+1}$
$=\dfrac{x+1}{x(x-1)}.\dfrac{(x-1)^2}{x+1}$
$=\dfrac{x-1}{x}$
$b)N=(\dfrac{x}{x+3}-\dfrac{x^2+9}{x^2-9}):(\dfrac{3x+1}{x^2-3x}-\dfrac{1}{x})$
$=[\dfrac{x(x-3)}{(x+3)(x-3)}-\dfrac{x^2+9}{(x+3)(x-3)}]:[\dfrac{3x+1}{x(x-3)}-\dfrac{x-3}{x(x-3)}$
$=\dfrac{x^2-3x-x^2-9}{(x+3)(x-3)}:\dfrac{3x+1-x+3}{x(x-3)}$
$=\dfrac{-3x-9}{(x+3)(x-3)}:\dfrac{2x+4}{x(x-3)}$
$=\dfrac{-3}{x-3}.\dfrac{x(x-3)}{2(x+2)}$
$=\dfrac{-3x}{2x+4}$
Bài 6:
$a)9x-3=0$
$⇒9x=3$
$⇒x=\dfrac{1}{3}$
Vậy $x=\dfrac{1}{3}$
$b)2x-\sqrt{5}=3$
$⇒2x=\sqrt{5}+3$
$⇒x=\dfrac{\sqrt{5}+3}{2}$
Vậy $x=\dfrac{\sqrt{5}+3}{2}$
$c)\dfrac{12x+4}{3}=-2$
$⇒12x+4=-6$
$⇒12x=-10$
$⇒x=-\dfrac{5}{6}$
Vậy $x=-\dfrac{5}{6}$
$d)2x^2=x$
$⇒2x^2-x=0$
$⇒x(2x-1)=0$
$⇒\left[ \begin{array}{l}x=0\\2x-1=0\end{array} \right. ⇒\left[ \begin{array}{l}x=0\\x=\dfrac{1}{2}\end{array} \right.$
Vậy $x∈\{0;\dfrac{1}{2}\}$
$e)x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}$
$⇒\dfrac{24x}{24}-\dfrac{4(5x+2)}{24}=\dfrac{6(7-3x)}{24}$
$⇒24x-20x-8=42-18x$
$⇒4x-8=-18x+42$
$⇒22x=50$
$⇒x=\dfrac{25}{11}$
Vậy $x=\dfrac{25}{11}$
$f)x^3-7x+6=0$
$⇒x^3-2x^2+2x^2-4x-3x+6=0$
$⇒x^2(x-2)+2x(x-2)-3(x-2)=0$
$⇒(x-2)(x^2+2x-3)=0$
$⇒(x-2)(x^2+3x-x-3)=0$
$⇒(x-2)[x(x+3)-(x+3)]=0$
$⇒(x-2)(x+3)(x-1)=0$
$⇒\left[ \begin{array}{l}x-2=0\\x+3=0\\x-1=0\end{array} \right. ⇒\left[ \begin{array}{l}x=2\\x=-3\\x=1\end{array} \right.$
Vậy $x∈\{-3;1;2\}$
