a, đk : x khác 2 ; -2 ; -3
\(A=\left(\dfrac{x+x+2}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x^2-4}{x+3}=\dfrac{2x+2}{x+3}\)
Vậy ta có đpcm
b, \(A=\dfrac{2x+2}{x+3}=\dfrac{2\left(x+3\right)-4}{x+3}=2-\dfrac{4}{x+3}\Rightarrow x+3\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| x + 3 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | -2 | -4 | -1 | -5 | 1 | -7 |
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a.A=\(\left(\dfrac{x}{x^2-4}+\dfrac{1\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x^2-4}{x+3}\)
A=\(\dfrac{x+x+2}{x^2-4}.\dfrac{x^2-4}{x+3}\)
A=\(\dfrac{2x+2}{x+3}\) ( đfcm)
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