\(\frac{x^3+x^2-x-1}{x^3+2x-5}\)
\(\Leftrightarrow\frac{x^3+x^2-x-1}{x^3+2x-5}=0\)
\(\Leftrightarrow\frac{x^2\left(x+1\right)-\left(x+1\right)}{x^3+2x-5}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x^2-1^2\right)}{x^3+2x-5}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+1\right)\left(x+1\right)}{x^3+2x-5}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2\left(x-1\right)}{x^3+2x-5}=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x-1\right)=0\)
\(\Leftrightarrow x=\pm1\)
Vậy \(x\in\left\{\pm1\right\}\)
\(\frac{x^3+x^2-x-1}{x^3+2x-5}=\frac{x^2\left(x+1\right)-\left(x+1\right)}{x^3+2x-5}\)
\(=\frac{\left(x+1\right)\left(x^2-1\right)}{x^3+2x-5}\)
Để \(\frac{x^3+x^2-x-1}{x^3+2x-5}=0\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\left(x^3+2x-5\ne0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2=1\end{cases}\Leftrightarrow x=\pm}\)
Vậy x={-1;1}
ĐKXĐ: x3 + 2x - 5 \(\ne\) 0
Khi đó x3 + x2 - x - 1 = 0
<=> x2(x + 1) - (x + 1) = 0
<=> (x2 - 1)(x + 1) = 0
<=> (x - 1)(x+1)2 = 0 <=> \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}\left(TM\right)}\)
Vậy ...
để giá trị của phân thức trên bằng 0 thì x^3+x^2-x-1=0
ta có:X^3+x^2-x-1=0
(x^3+x^2)-(x+1)=0
x^2(x+1)-(x+1)=0
(x^2-1)(x+1)=0
(x-1)(x+1)^2=0
suy ra x-1=0 hoặc (x+1)^2=0
xét các th:
*x-1=0 suy ra x=1
*(x+1)^2=0 suy ra x+1=0 suy ra x=-1
vậy để phân thức =0 thì x=1 hoặc x=-1
