Question

[✎﹏トラン⋮ Hannie ッ]

Xác định các giá trị của \(m\) để có số \(x< 0\) sao cho:

\(m=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)

Answer 1

[Trần Tuấn Hoàng]

-ĐKXĐ: \(x\ne\pm1\)

\(\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)

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

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

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

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

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

\(\Rightarrow m=\dfrac{x}{x^2+1}\)

-Khi \(x< 0\), mà \(x^2+1>0\forall x\)

\(\Rightarrow m=\dfrac{x}{x^2+1}< 0\).

\(\Rightarrow m< 0\)

-Vậy khi \(m< 0\) và \(m\ne\dfrac{-1}{2}\) thì \(x< 0\)