Question

[Arceus Official]

Cho B = \(\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)

a)Rút gọn B

b)Tìm giá trị của x để B<0

c)Tìm giá trị của B để x thoã mãn : \(x-4=5\)

Answer 1

[Minh Nguyen]

\(ĐKXĐ:x\ne\pm1\)

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 \(x-4=5\Leftrightarrow B=-656\)