Question

Giải pt:\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)

Answer 1

điều kiện xác định : \(x\ne-2\)

ta có : \(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)

\(\Leftrightarrow\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}-\dfrac{5}{x^2-2x+4}=0\)

\(\Leftrightarrow\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5}{x^2-2x+4}=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-4x+8}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5x+10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-4x+8-2x^2-16-5x-10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

\(\Leftrightarrow\dfrac{-9x-18}{\left(x+2\right)\left(x^2-2x+4\right)}=0\Leftrightarrow-9x-18=0\)

\(\Leftrightarrow-9x=18\Leftrightarrow x=-2\left(loại\right)\)

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Answer 2

Giải:

\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\) (1)

ĐKXĐ: \(x\ne-2\)

\(\left(1\right)\Leftrightarrow\dfrac{2\left(x^2-2x+4\right)}{x^3+8}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5\left(x+2\right)}{x^3+8}\)

\(\Rightarrow2\left(x^2-2x+4\right)-2x^2+16=5\left(x+2\right)\)

\(\Rightarrow2x^2-4x+8-2x^2+16=5x+10\)

\(\Rightarrow-4x-5x=10-8-16\)

\(\Rightarrow-9x=-14\)

\(\Rightarrow x=-\dfrac{14}{-9}=\dfrac{14}{9}\) (thoả mãn ĐKXĐ)

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