Giải giúp mik với ak

Giải thích các bước giải:

a.ĐKXĐ: $x\ne0$

Ta có:

$\dfrac{3x^2+7x-10}x=0$ 

$\to 3x^2+7x-10=0$

$\to (x-1)(3x+10)=0$

$\to x\in\{1, -\dfrac{10}3\}$

b.Ta có:

$\dfrac{4x-17}{2x^2+1}=0$

$\to 4x-17=0$

$\to 4x=17$
$\to x=\dfrac{17}4$

c.ĐKXĐ: $x\ne -2$

Ta có:

$\dfrac{(x^2+2x)-(3x+6)}{x+2}=0$

$\to \dfrac{x(x+2)-3(x+2)}{x+2}=0$

$\to x-3=0$

$\to x=3$

d.ĐKXĐ: $x\ne 3$

Ta có:
$\dfrac{x^2-x-6}{x-3}=0$

$\to \dfrac{x^2-3x+2x-6}{x-3}=0$

$\to \dfrac{x(x-3)+2(x-3)}{x-3}=0$

$\to x+2=0$

$\to x=-2$

e.ĐKXĐ: $x\ne-5$

Ta có:

$\dfrac{2x-5}{x+5}=3$

$\to 2x-5=3x+15$

$\to x=-20$

f.ĐKXĐ: $x\ne-\dfrac23$

Ta có:

$\dfrac5{3x+2}=2x-1$

$\to 5=(3x+2)(2x-1)$

$\to 5=6x^2+x-2$

$\to 6x^2+x-7=0$

$\to (x-1)(6x+7)=0$
$\to x\in\{1, -\dfrac76\}$

g.ĐKXĐ: $x\ne 0$

$\dfrac{x^2-6}x=x+\dfrac32$

$\to x-\dfrac6x=x+\dfrac32$

$\to -\dfrac6x=\dfrac32$

$\to x=-4$

h.ĐKXĐ: $x\ne 2$

$\dfrac4{x-2}-x+2=0$

$\to 4-(x-2)^2=0$

$\to 4-x^2+4x-4=0$

$\to x^2-4x=0$

$\to x(x-4)=0$
$\to x\in\{0,4\}$