Dạ giúp tớ bài này với ạ

Đề bài: giải phương trình(dùng cách tách)

Giải thích các bước giải:

c.ĐKXĐ: $x\ne -2, x\ne -1$

Ta có:

$\dfrac{2x+1}{x+2}-\dfrac{x+3}{x+1}+1=0$

$\to \left(2x+1\right)\left(x+1\right)-\left(x+3\right)\left(x+2\right)+\left(x+2\right)\left(x+1\right)=0$

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

$\to (2x+3)(x-1)=0$

$\to x\in\{1, \dfrac{-3}2\}$

d.ĐKXĐ: $x\ne 0, x\ne 18$

Ta có:

$\dfrac1x+\dfrac1{18-x}=\dfrac14$

$\to 4(18-x)+4x=x(18-x)$

$\to 72=18x-x^2$

$\to x^2-18x+72=0$

$\to (x-6)(x-12)=0$

$\to x\in\{6, 12\}$

e.ĐKXĐ $x\ne 0, x\ne -3$

Ta có:

$\dfrac{480}x-\dfrac{480}{x+3}=8$

$\to 480\left(x+3\right)-480x=8x\left(x+3\right)$

$\to 1440=8x^2+24x$

$\to 8x^2+24x-1440=0$

$\to 8(x-12)(x+15)=0$

$\to x\in\{12, -15\}$

f.ĐKXĐ: $x\ne -1, x\ne 2$

Ta có:

$\dfrac3{x+1}-\dfrac1{x-2}=\dfrac9{(x+1)(x-2)}$

$\to \dfrac{2x-7}{(x+1)(x-2)}=\dfrac9{(x+1)(x-2)}$

$\to 2x-7=9$

$\to 2x=16$

$\to x=8$

a.ĐKXĐ: $x\ne 0, x\ne 3$

Ta có:

$\dfrac{x+3}x=\dfrac{x+9}{x-3}$

$\to (x+3)(x-3)=x(x+9)$

$\to x^2-9=x^2+9x$

$\to 9x=-9$

$\to x=-1$

b.ĐKXĐ: $x\ne 1, x\ne 0$

Ta có:

$\dfrac{x}{x-1}+\dfrac{x+1}x=2$

$\to x^2+(x+1)(x-1)=2x(x-1)$

$\to 2x^2-1=2x^2-2x$

$\to 2x=1$

$\to x=\dfrac12$