ĐK: \(x \neq - 1 , \textrm{ }\textrm{ } x \neq - 2\)

\(\frac{2}{x + 1} - \frac{3}{x + 2} = \frac{1}{3 x + 3}\) \(\Leftrightarrow \frac{2}{x + 1} - \frac{3}{x + 2} = \frac{1}{3 \left(\right. x + 1 \left.\right)}\)

\(\Leftrightarrow 6 \left(\right. x + 2 \left.\right) - 9 \left(\right. x + 1 \left.\right) = x + 2\)

\(\Leftrightarrow 6 x + 12 - 9 x - 9 = x + 2\)

\(\Leftrightarrow - 3 x + 3 = x + 2\)

\(\Leftrightarrow - 4 x = - 1\)

\(\Leftrightarrow x = \frac{1}{4}\)

Vậy \(x = \frac{1}{4}\).

ĐKXĐ: x+1≠0

=> x≠-1

x+2≠0

=> x≠-2

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

\(\frac{6\left(x+2\right)}{3\left(x+1\right)\left(x+2\right)}-\frac{9\left(x+1\right)}{3\left(x+1\right)\left(x+2\right)}=\frac{\left(x+2\right)}{3\left(x+1\right)\left(x+2\right)}\)

\(6\left(x+2\right)-9\left(x+1\right)=x+2\)

\(6x+12-9x-9=x+2\)

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

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

\(-4x=-1\)

\(x=\frac14\)

vậy x=\(\frac14\)

ĐKXĐ:

+) \(x+1\ne0\Rightarrow x\ne-1\)

+) \(x+2\ne0\Rightarrow x\ne-2\)

+) \(3x+3\ne0\Rightarrow3\left(x+1\right)\Rightarrow x\ne-1\)

Do đó, ĐKXĐ là \(x\ne-1\) và \(x\ne-2\)

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

\(\Leftrightarrow\frac{2}{x+1}-\frac{3}{x+2}=\frac{1}{3\left(x+1\right)}\)

\(\Leftrightarrow\frac{2.3.\left(x+2\right)}{3.\left(x+1\right)\left(x+2\right)}-\frac{3.3.\left(x+1\right)}{3.\left(x+1\right)\left(x+2\right)}=\frac{x+2}{3.\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow6.\left(x+2\right)-9.\left(x+1\right)=x+2\)

\(\Leftrightarrow6x+12-9x-9=x+2\)

\(\Leftrightarrow-3x+3=x+2\)

\(\Leftrightarrow-3x-x=2-3\)

\(\Leftrightarrow-4x=-1\)

\(\Leftrightarrow x=\frac14\) (TMĐK)

Vậy nghiệm pt là \(\frac14\)

Điều kiện: x ≠ -1, x ≠ -2
2/(x + 1) - 3/(x + 2) = 1/(3x + 3)
2/(x + 1) - 3/(x + 2) = 1/[3(x + 1)]
Nhân hai vế với 3(x + 1)(x + 2):
6(x + 2) - 9(x + 1) = x + 2
6x + 12 - 9x - 9 = x + 2
-3x + 3 = x + 2
-4x = -1
x = 1/4
Vậy x = 1/4, giá trị này thỏa mãn điều kiện nên là nghiệm của phương trình

ĐKXĐ: $x\ne-1,\ x\ne-2,\ x\ne-3$
$\dfrac{2}{x+1}-\dfrac{3}{x+2}=\dfrac{1}{3x+3}$
$\dfrac{2}{x+1}-\dfrac{3}{x+2}=\dfrac{1}{3(x+1)}$
$\dfrac{6}{3(x+1)}-\dfrac{9}{3(x+2)}=\dfrac{1}{3(x+1)}$
$\dfrac{5}{3(x+1)}=\dfrac{9}{3(x+2)}$
$\dfrac{5}{x+1}=\dfrac{9}{x+2}$
$5(x+2)=9(x+1)$
$5x+10=9x+9$
$4x=1$
$x=\dfrac14$