nhanh giúp mình nhé

To solve the given equation, we can start by expressing each fraction in partial fractions. We have: \[\frac{1}{x^2+9x+20} = \frac{1}{(x+4)(x+5)},\] \[\frac{1}{x^2+11x+30} = \frac{1}{(x+5)(x+6)},\] \[\frac{1}{x^2+13x+42} = \frac{1}{(x+6)(x+7)}.\] So the given equation becomes: \[\frac{1}{(x+4)(x+5)}+\frac{1}{(x+5)(x+6)}+\frac{1}{(x+6)(x+7)}=\frac{1}{18}.\] We can then find a common denominator and combine the fractions on the left-hand side: \[\frac{(x + 6) + (x + 4) + (x + 5)}{(x + 4)( x + 5)( x + 6)( x + 7)} = \frac{1}{18},\] \[ \Rightarrow \frac{3(x^2 + 12 x + 15)}{( x^2 + 11 x + 30 )( x^2 + 13 x + 42 )} = \frac{1}{18},\] \[ \Rightarrow \frac{x^2 + 12 x + 15 }{( x^2 + 11 x + 30 )( x^2 + 13 x +42 )} = \frac{1}{54},\] \[ \Rightarrow (54)x^2+(648)x+(810)= (18)x^4+(204)x^3+(540)x^2+(702)x+(1260),\] \[0=18(x-3)^2(x-2)^2.\] From this, we get two solutions for \( x: -13\) and \(2.\) Therefore, the solution to the given equation is \( [-13,~2].\)