Câu 30 hoàn chỉnh là x^2+3x+2

Giải thích các bước giải:

$\begin{aligned}
& \quad \dfrac{x+1}{x+2} : \dfrac{x+2}{x+3} \cdot \dfrac{x+3}{x+1} \\
& = \dfrac{x+1}{x+2} \cdot \dfrac{x+3}{x+2} \cdot \dfrac{x+3}{x+1} \\
& = \dfrac{(x+1)(x+3)(x+3)}{(x+2)(x+2)(x+1)} \\
& = \dfrac{(x+3)^2}{(x+2)^2}
\end{aligned}$

 $\begin{aligned}
& \quad \dfrac{x+1}{x+2} : \dfrac{x+2}{x+3} : \dfrac{x+3}{x+1} \\
& = \dfrac{x+1}{x+2} \cdot \dfrac{x+3}{x+2} \cdot \dfrac{x+1}{x+3} \\
& = \dfrac{(x+1)(x+3)(x+1)}{(x+2)(x+2)(x+3)} \\
& = \dfrac{(x+1)^2}{(x+2)^2}
\end{aligned}$

$\begin{aligned}
 & \quad \dfrac{x^2+x}{x+2} : \dfrac{x+3}{x+2} \cdot \dfrac{x+3}{x+1} \\
& = \dfrac{x(x+1)}{x+2} \cdot \dfrac{x+2}{x+3} \cdot \dfrac{x+3}{x+1} \\
& = \dfrac{x(x+1)(x+2)(x+3)}{(x+2)(x+3)(x+1)} \\
& = x
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{x+1}{x+2} : \left( \dfrac{x+2}{x+3} : \dfrac{x+3}{x+1} \right) \\
& = \dfrac{x+1}{x+2} : \left( \dfrac{x+2}{x+3} \cdot \dfrac{x+1}{x+3} \right) \\
& = \dfrac{x+1}{x+2} \cdot \dfrac{(x+3)^2}{(x+2)(x+1)} \\
& = \dfrac{(x+1)(x+3)^2}{(x+2)^2(x+1)} \\
& = \dfrac{(x+3)^2}{(x+2)^2}
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{x-2}{x+1} \cdot \dfrac{x^2-2x-3}{x^2-5x+6} \\
& = \dfrac{x-2}{x+1} \cdot \dfrac{(x+1)(x-3)}{(x-2)(x-3)} \\
& = \dfrac{(x-2)(x+1)(x-3)}{(x+1)(x-2)(x-3)} \\
& = 1
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{x+1}{x^2-2x-8} \cdot \dfrac{4-x}{x^2+x} \\
& = \dfrac{x+1}{(x-4)(x+2)} \cdot \dfrac{-(x-4)}{x(x+1)} \\
& = \dfrac{-(x+1)(x-4)}{x(x-4)(x+2)(x+1)} \\
& = \dfrac{-1}{x(x+2)}
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{x^2+2}{4x+24} \cdot \dfrac{x^2-36}{x^2+x-2} \\
& = \dfrac{x^2+2}{4(x+6)} \cdot \dfrac{(x-6)(x+6)}{(x+2)(x-1)} \\
& = \dfrac{(x^2+2)(x-6)(x+6)}{4(x+6)(x+2)(x-1)} \\
& = \dfrac{(x^2+2)(x-6)}{4(x+2)(x-1)}
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{x^3-8}{5x+20} \cdot \dfrac{x^2+4x}{x^2+2x+4} \\
& = \dfrac{(x-2)(x^2+2x+4)}{5(x+4)} \cdot \dfrac{x(x+4)}{x^2+2x+4} \\
& = \dfrac{x(x-2)(x+4)(x^2+2x+4)}{5(x+4)(x^2+2x+4)} \\
& = \dfrac{x(x-2)}{5}
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{(x-3)^3}{3x^2} : \dfrac{x^2-6x+9}{6x} \\
& = \dfrac{(x-3)^3}{3x^2} \cdot \dfrac{6x}{(x-3)^2} \\
& = \dfrac{6x(x-3)^3}{3x^2(x-3)^2} \\
& = \dfrac{2(x-3)}{x}
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{x^2+x}{5x^2-10x+5} : \dfrac{3x+3}{5x-5} \\
& = \dfrac{x(x+1)}{5(x-1)^2} : \dfrac{3(x+1)}{5(x-1)} \\
& = \dfrac{x(x+1)}{5(x-1)^2} \cdot \dfrac{5(x-1)}{3(x+1)} \\
& = \dfrac{5x(x+1)(x-1)}{15(x-1)^2(x+1)} \\
& = \dfrac{x}{3(x-1)}
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{6x-3}{5x^2+x} : \dfrac{1-8x^3}{25x^2+10x+1} \\
& = \dfrac{3(2x-1)}{x(5x+1)} : \dfrac{(1-2x)(1+2x+4x^2)}{(5x+1)^2} \\
& = \dfrac{3(2x-1)}{x(5x+1)} \cdot \dfrac{(5x+1)^2}{-(2x-1)(4x^2+2x+1)} \\
& = \dfrac{3(2x-1)(5x+1)^2}{-x(5x+1)(2x-1)(4x^2+2x+1)} \\
& = \dfrac{-3(5x+1)}{x(4x^2+2x+1)}
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{x^2+2x+1}{x^3-1} : \dfrac{x+1}{2x^2+2x+2} \\
& = \dfrac{(x+1)^2}{(x-1)(x^2+x+1)} : \dfrac{x+1}{2(x^2+x+1)} \\
& = \dfrac{(x+1)^2}{(x-1)(x^2+x+1)} \cdot \dfrac{2(x^2+x+1)}{x+1} \\
& = \dfrac{2(x+1)^2(x^2+x+1)}{(x-1)(x^2+x+1)(x+1)} \\
& = \dfrac{2(x+1)}{x-1}
\end{aligned}$

$\begin{aligned}
& \quad \dfrac{x+2}{x^2-5x+6} : \dfrac{x^2+3x+2}{x-2} \\
& = \dfrac{x+2}{(x-2)(x-3)} : \dfrac{(x+1)(x+2)}{x-2} \\
& = \dfrac{x+2}{(x-2)(x-3)} \cdot \dfrac{x-2}{(x+1)(x+2)} \\
& = \dfrac{(x+2)(x-2)}{(x-2)(x-3)(x+1)(x+2)} \\
& = \dfrac{1}{(x-3)(x+1)} \\
& = \dfrac{1}{x^2-2x-3}
\end{aligned}$