Question

Làm câu 1a giúp mình cái nhé!

Answer 1

a: Ta có: \(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{1}{x+2}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{1}{x+2}:\left(\dfrac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=1+\dfrac{1}{x+2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{6}\)

\(=1+\dfrac{x-2}{6}\)

\(=\dfrac{x+4}{6}\)