Giúp với mọi người ơi

`a)` $\frac{x-1}{2x}+\frac{2x+1}{3x}+\frac{1-5x}{6x}$

 `MSC`: `6x`

$= \frac{x-1}{2x} \cdot \frac{3}{3} + \frac{2x+1}{3x} \cdot \frac{2}{2} + \frac{1-5x}{6x}$

$=\frac{3(x-1)}{6x}+\frac{2(2x+1)}{6x}+\frac{1-5x}{6x}$

$=\frac{3x-3+4x+2+1-5x}{6x}$

$=\frac{2x}{6x}$

$=\frac{1}{3}$

`b)`$\frac{1}{x-y}+\frac{2}{x+y}+\frac{3}{y^2-x^2}$

`MSC`: $(x-y)(x+y)$.

$=\frac{1}{x-y} \cdot \frac{x+y}{x+y} + \frac{2}{x+y} \cdot \frac{x-y}{x-y} + \frac{3}{(y-x)(y+x)}$

$=\frac{x+y}{(x-y)(x+y)}+\frac{2(x-y)}{(x-y)(x+y)}+\frac{-3}{(x-y)(x+y)}$

$=\frac{x+y+2x-2y-3}{(x-y)(x+y)}$

$=\frac{3x-y-3}{(x-y)(x+y)}$

`c)`$\frac{1}{a-b-c}+\frac{1}{b-c-a}+\frac{1}{c-a-b}$

$=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

$=\frac{yz+xz+xy}{xyz}$

Vì $x+y+z = 0$, nên ta có:

$xy + yz + zx = 0$ 

Do đó, biểu thức trên bằng `0`.

`a,`

`(x-1)/(2x)+(2x+1)/(3x)+(1-5x)/(6x)`

`= (3(x-1)+2(2x+1)+1-5x)/(6x)` 

`= (3x-3+4x+2+1-5x)/(6x)`

`= (2x)/(6x)`

`= 1/3`

`b,`

`1/(x-y)+2/(x+y)+3/(y^2-x^2)`

`= 1/(x-y)+2(x+y)-3/((x-y)(x+y)`

`= (x+y+2(x-y)-3)/((x-y)(x+y)`

`= (x+y+2x-2y-3)/((x-y)(x+y)`

`= (3x-y-3)/((x-y)(x+y)`

`c,`

`1/((a-b)(b-c))+1/((b-c)(c-a))+1/((c-a)(a-b))`

`= (c-a+a-b+b-c)/((a-b)(b-c)(c-a))`

`= 0/((a-b)(b-c)(c-a))`

`= 0`

`#TD`