a, Thay x = 1 vào A ta được
\(A=\dfrac{2}{\left(x+1\right)^2}\Rightarrow A=\dfrac{2}{2^2}=\dfrac{1}{2}\)
b, Với x khác 0 ; -1
\(B=\dfrac{x^2-x^2+1}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\)
\(P=\dfrac{2}{\left(x+1\right)^2}:\dfrac{1}{x\left(x+1\right)}=\dfrac{2x}{x+1}\)
Ta có \(P=\dfrac{2x}{x+1}=-3\Leftrightarrow2x=-3x-3\Leftrightarrow x=-\dfrac{3}{5}\)
c, \(P=\dfrac{2x}{x+1}=\dfrac{2\left(x+1\right)-2}{x+1}=2-\dfrac{2}{x+1}\Rightarrow x+1\inƯ\left(-2\right)=\left\{\pm1;\pm2\right\}\)
| x+1 | 1 | -1 | 2 | -2 |
| x | 0(ktm) | -2 | 1 | -3 |
Đúng 0
Bình luận (0)
