\(ĐKXĐ:x\ne\pm1\)
a) \(B=\left(\frac{1-x^3}{1-x}-x\right)\div\frac{1-x^2}{1-x-x^2+x^3}\)
\(\Leftrightarrow B=\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right):\left(\frac{\left(1-x\right)\left(1+x\right)}{\left(x-1\right)^2\left(x+1\right)}\right)\)
\(\Leftrightarrow B=\left(1+x+x^2-x\right):\left(\frac{-1}{x-1}\right)\)
\(\Leftrightarrow B=-\left(x^2+1\right).\left(x-1\right)\)
\(\Leftrightarrow B=-x^3+x^2-x+1\)
b) Để B < 0
\(\Leftrightarrow-x^3+x^2-x+1< 0\)
\(\Leftrightarrow-\left(x^2+1\right)\left(x-1\right)< 0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)>0\)
TH1 : \(\hept{\begin{cases}x^2+1>0\left(tm\right)\\x-1>0\end{cases}\Leftrightarrow x>1}\)
TH2 : \(\hept{\begin{cases}x^2+1< 0\left(ktm\right)\\x-1< 0\end{cases}}\Leftrightarrow x\in\varnothing\)
Vậy để \(B< 0\Leftrightarrow x>1\)
c) Khi \(x-4=5\)
\(\Leftrightarrow x=9\)
\(\Leftrightarrow B=-\left(9^3\right)+9^2-9+1\)
\(\Leftrightarrow B=-729+81-9+1\)
\(\Leftrightarrow B=-656\)
Vậy khi \(x-4=5\Leftrightarrow B=-656\)
