Môn Toán Lớp 8: thực hiện phép tính 1/x+3+1/(x+3)(x+2)+1/(x+2)(4x+7)

\[\begin{array}{l}
\# andy\\
\dfrac{1}{{x + 3}} + \dfrac{1}{{\left( {x + 3} \right)\left( {x + 2} \right)}} + \dfrac{1}{{\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{1.\left( {x + 2} \right)\left( {4x + 7} \right)}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}} + \dfrac{{1.\left( {4x + 7} \right)}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}} + \dfrac{{1.\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{4{x^2} + 7x + 8x + 14}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}} + \dfrac{{4x + 7}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}} + \dfrac{{x + 3}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{4{x^2} + 7x + 8x + 14 + 4x + 7 + x + 3}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{4{x^2} + 20x + 24}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{4\left( {{x^2} + 5x + 6} \right)}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{4\left( {{x^2} + 3x + 2x + 6} \right)}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{4\left[ {\left( {{x^2} + 2x} \right) + \left( {3x + 6} \right)} \right]}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{4\left[ {x\left( {x + 2} \right) + 3\left( {x + 2} \right)} \right]}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{{4\left( {x + 3} \right)\left( {x + 2} \right)}}{{\left( {x + 3} \right)\left( {x + 2} \right)\left( {4x + 7} \right)}}\\
 = \dfrac{4}{{4x + 7}}
\end{array}\]