Mọi người chỉ giúp em phần e,L ạ.

Giải thích các bước giải:

e.ĐKXĐ: $x\ne 2, 4$
Ta có:
$\dfrac{x-3}{x-2}-\dfrac{x-2}{x-4}=1\dfrac{5}{21}$
$\to \dfrac{x-2-1}{x-2}-\dfrac{x-4+2}{x-4}=\dfrac{26}{21}$
$\to 1-\dfrac{1}{x-2}-(1+\dfrac{2}{x-4})=\dfrac{26}{21}$
$\to -\dfrac{1}{x-2}-\dfrac{2}{x-4}=\dfrac{26}{21}$

$\to -21\left(x-4\right)-42\left(x-2\right)=26\left(x-2\right)\left(x-4\right)$

$\to -63x+168=26x^2-156x+208$

$\to 26x^2-93x+40=0$

$\to (26x^2-13x)-(80x-40)=0$

$\to 13x(2x-1)-40(2x-1)=0$

$\to (13x-40)(2x-1)=0$

$\to 13x-40=0\to x=\dfrac{40}{13}$

Hoặc $2x-1=0\to 2x=1\to x=\dfrac12$

l.Ta có:

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

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

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

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

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

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

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

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

$\to \dfrac{x+1}{x-1}=3$

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

$\to x+1=3x-3$

$\to 2x=4$

$\to x=2$