Giải đáp: \(\dfrac{{ – 5x – 4}}{{x\left( {x + 4} \right)}}\) Lời giải và giải thích chi tiết: \(\begin{array}{l}DK:x \ne \left\{ { – 4;0} \right\}\\\dfrac{x}{{x + 4}} + \dfrac{{x – 1}}{x} – 2\\ = \dfrac{{{x^2} + \left( {x – 1} \right)\left( {x + 4} \right) – 2x\left( {x + 4} \right)}}{{x\left( {x + 4} \right)}}\\ = \dfrac{{{x^2} + {x^2} + 3x – 4 – 2{x^2} – 8x}}{{x\left( {x + 4} \right)}}\\ = \dfrac{{ – 5x – 4}}{{x\left( {x + 4} \right)}}\end{array}\)
~ Bạn tham khảo ~ x/[x+4] + [x-1]/x – 2 ĐKXĐ : x\ne0;-4 = [x^2]/[x(x+4)] + [(x-1)(x+4)]/[x(x+4)] – [2x(x+4)]/[x(x+4)] = [x^2+x^2+4x-x-4-2x^2-8x]/[x(x+4)] = [-5x-4]/[x(x+4)]
DK:x \ne \left\{ { – 4;0} \right\}\\
\dfrac{x}{{x + 4}} + \dfrac{{x – 1}}{x} – 2\\
= \dfrac{{{x^2} + \left( {x – 1} \right)\left( {x + 4} \right) – 2x\left( {x + 4} \right)}}{{x\left( {x + 4} \right)}}\\
= \dfrac{{{x^2} + {x^2} + 3x – 4 – 2{x^2} – 8x}}{{x\left( {x + 4} \right)}}\\
= \dfrac{{ – 5x – 4}}{{x\left( {x + 4} \right)}}
\end{array}\)