a, \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}=\frac{x+1}{2\left(x+3\right)}+\frac{3x+2}{x\left(x+3\right)}\)

\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{6x+4}{2x\left(x+3\right)}=\frac{x^2+7x+4}{2x\left(x+3\right)}\)

b, Sua de :  \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)

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

\(\dfrac{x^2-2x-8}{2x^2+9x+10}\)

\(=\dfrac{x^2-4x+2x-8}{2x^2+4x+5x+10}\)

\(=\dfrac{\left(x-4\right)\left(x+2\right)}{\left(x+2\right)\left(2x+5\right)}\)

\(=\dfrac{x-4}{2x+5}\)

Các bạn giúp mình với nhé thanks 

\(\dfrac{x^2-2x-8}{2x^2+9x+10}\)

\(=\dfrac{\left(x-4\right)\left(x+2\right)}{2x^2+4x+5x+10}\)

\(=\dfrac{\left(x-4\right)\left(x+2\right)}{\left(x+2\right)\left(2x+5\right)}\)

\(=\dfrac{x-4}{2x+5}\)

Ta có: \(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

\(\dfrac{x+10}{x-2}+\dfrac{x-18}{x+2}+\dfrac{x+2}{x^2-4}=\dfrac{\left(x+10\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-18\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+12x+20+x^2-16x-36+x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2-3x-14}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x^2+4x\right)-\left(7x+14\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x\left(x+2\right)-7\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x-7\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-7}{x-2}\)

Cái này đã là nhân tử rồi mà bạn

Điều kiện xác định của phân thức: x ≠ -10, x ≠ 10

Vậy giá trị P =10 với mọi x ≠ ± 10