\(B=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)   \(ĐKXĐ:x\ne\pm1\)

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

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

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

\(B=\frac{x^2+1}{1-x}\)

vậy \(B=\frac{x^2+1}{1-x}\)

b) \(x=-1\frac{2}{3}\)

\(x=\frac{-5}{3}\)

khi đó \(B=\frac{\left(\frac{-5}{3}\right)^2+1}{1+\frac{5}{3}}\)

\(B=\frac{\frac{25}{9}+1}{\frac{8}{3}}\)

\(B=\frac{34}{9}:\frac{8}{3}\)

\(B=\frac{17}{12}\)

vậy \(B=\frac{17}{12}\) khi \(x=-1\frac{2}{3}\)

c) \(B< 0\Leftrightarrow\frac{x^2+1}{1-x}< 0\)

\(\Leftrightarrow\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+1< 0\\1-x>0\end{cases}}\)

đến đây bạn giải tiếp 

=> 2x+2/x - 5 > -1

=> 2x+2/x > -1 + 5 = 4

=> 2x+2 > 4x

=> 2x+2-4x > 0

=> 2-2x > 0

=> 2x < 2

=> x < 2 : 2 = 1

Vậy x thuộc Z và x < 1

Tk mk nha 

Hóa học 8 đấy các bạn ạk

Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)

Đặt \(y=x^2+6x+5\), ta có: 

\(y\left(y+3\right)-40=0\)

\(\Leftrightarrow\left(y-5\right)\left(y+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=5\\y=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)

Ta có:  \(\frac{-x^2-1}{x}< -1\)

\(\Rightarrow\frac{-x^2-1}{x}+1< 0\Rightarrow\frac{-x^2+x-1}{x}< 0\)

Ta thấy \(x^2-x+1>0\forall x\Rightarrow-x^2+x-1< 0\)

Vậy để \(\frac{-x^2+x-1}{x}< 0\) thì \(x>0\)