1) Tìm x để phân thức sau bằng không:
\(\dfrac{x^3+x^2-x-1}{x^3+2x-3}\)
2) Tính tổng:
\(B=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
Làm theo hướng dẫn: \(\dfrac{1}{k\left(k+1\right)\left(k+2\right)}=\dfrac{1}{2}\left(\dfrac{1}{k}+\dfrac{1}{k+2}\right)-\dfrac{1}{k+1}\)
Bài 1 :
Để \(\dfrac{x^3+x^2-x-1}{x^3+2x-3}=0\) thì \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy,.........
