Question

thực hiện phép tính:

(1/x + 2) + (5/2x^2 + 3x - 2 )

Answer 1

ĐKXĐ: \(x\notin\left\{-2;\dfrac{1}{2}\right\}\)

\(\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}\)

\(=\dfrac{1}{x+2}+\dfrac{5}{\left(x+2\right)\left(2x-1\right)}\)

\(=\dfrac{2x-1+5}{\left(x+2\right)\left(2x-1\right)}=\dfrac{2x+4}{\left(x+2\right)\left(2x-1\right)}\)

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

Answer 2

\(\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}\)

\(=\dfrac{1}{x+2}+\dfrac{5}{2x^2+4x-x-2}\)

\(=\dfrac{1}{x+2}+\dfrac{5}{\left(2x^2+4x\right)-\left(x+2\right)}\)

\(=\dfrac{1}{x+2}+\dfrac{5}{2x\left(x+2\right)-\left(x+2\right)}\)

\(=\dfrac{1}{x+2}+\dfrac{5}{\left(x+2\right)\left(2x-1\right)}\)   (1)

ĐKXĐ: \(x\ne2;x\ne\dfrac{1}{2}\)

(1) \(=\dfrac{2x-1}{\left(x+2\right)\left(2x-1\right)}+\dfrac{5}{\left(x+2\right)\left(2x-1\right)}\)

\(=\dfrac{2x-1+5}{\left(x+2\right)\left(2x-1\right)}\)

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

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

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