Question

\(\dfrac{12}{8+x^3}\)=1+\(\dfrac{1}{x+2}\)

Answer 1

ĐKXĐ : \(x\ne-2\)

PT \(\Leftrightarrow\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow12=\left(x+2\right)\left(x^2-2x+4\right)+x^2-2x+4\)

\(\Leftrightarrow12=x^3+8+x^2-2x+4\)

\(\Leftrightarrow x^3+8+x^2-2x+4-12=0\)

\(\Leftrightarrow x^3+x^2-2x=0\)

\(\Leftrightarrow x\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)\left(x+2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=1\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy ...

 

 

Answer 2

\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)

⇔ \(\dfrac{12}{\left(2+x\right)\left(4-2x+x^2\right)}=\dfrac{8+x^3}{\left(2+x\right)\left(4-2x+x^2\right)}+\dfrac{4-2x+x^2}{\left(2+x\right)\left(4-2x+x^2\right)}\)

⇔ 12 = 8 + \(x^3\) + 4 - 2x + \(x^2\)

⇔ 12 -  8 - \(x^3\) - 4 + 2x - \(x^2\)= 0 

⇔ -\(x^3\) + 2x - \(x^2\) = 0 

⇔ -x(\(x^2-2+x\) ) = 0 

⇔ -x(\(x^2+2x-x-2\) ) = 0 

⇔ -x[x(x+2)-(x+2)] = 0 

⇔ -x(x-1)(x+2) = 0 

⇔ -x = 0 hoặc x-1 = 0 hoặc x + 2 = 0 

⇔ x = 0 hoặc x = 1 hoặc x = -2 

Vậy phương trình có tập nghiệm S = {0;1;-2}