\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}}\)
A.\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}} = \dfrac{2}{{x + 3}}\)
B.\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}} = \dfrac{2(x-3)}{{x + 3}}\)
C.\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}} = \dfrac{(x-3)}{{x + 3}}\)
D.\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}} = \dfrac{2}{{x - 3}}\)
A.\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}} = \dfrac{2}{{x + 3}}\)
B.\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}} = \dfrac{2(x-3)}{{x + 3}}\)
C.\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}} = \dfrac{(x-3)}{{x + 3}}\)
D.\(\dfrac{{x + 1}}{{x + 3}} - \dfrac{{x - 1}}{{3 - x}} + \dfrac{{2x - 2{x^2}}}{{{x^2} - 9}} = \dfrac{2}{{x - 3}}\)
Loga
Hóa Học lớp 12
