1)3/x+3- 1/x-2= 5/2x+6

2)7/2x-3 + 1/2x-2=3/x-1

3)1/2x-1 +3/12x-8 = 2/3x-2

Giúp e vớiiiiii

Giải thích các bước giải:

1.ĐKXĐ: $x\ne -3, x\ne 2$

Ta có:

$\dfrac3{x+3}-\dfrac1{x-2}=\dfrac5{2x+6}$

$\to \dfrac3{x+3}-\dfrac1{x-2}=\dfrac5{2(x+3)}$

$\to \dfrac1{2(x+3)}=\dfrac1{x-2}$

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

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

$\to  x=-8$

2.ĐKXĐ: $x\ne 1, x\ne\dfrac32$

Ta có:

$\dfrac7{2x-3}+\dfrac1{2x-2}=\dfrac3{x-1}$

$\to \dfrac7{2x-3}+\dfrac1{2(x-1)}=\dfrac3{x-1}$

$\to \dfrac7{2x-3}=\dfrac5{2(x-1)}$

$\to 14x-14=10x-15$

$\to 4x=-1$

$\to x=-\dfrac14$

3.ĐKXĐ: $x\ne\dfrac12, x\ne\dfrac23$

Ta có:
$\dfrac1{2x-1}+\dfrac3{12x-8}=\dfrac2{3x-2}$

$\to \dfrac1{2x-1}+\dfrac3{4(3x-2)}=\dfrac2{3x-2}$

$\to \dfrac1{2x-1}=\dfrac2{3x-2}-\dfrac3{4(3x-2)}$

$\to\dfrac1{2x-1}=\dfrac5{4(3x-2)}$

$\to 12x-8=10x-5$

$\to 2x=3$

$\to x=\dfrac32$