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Question 64
4x+8x−3⋅2x2−12x+18x2−4\frac{4 x+8}{x-3} \cdot \frac{2 x^{2}-12 x+18}{x^{2}-4}x−34x+8⋅x2−42x2−12x+18
A) 8(x+2) (x−3) x2−4\frac{8(x+2) (x-3) }{x^{2}-4}x2−48(x+2) (x−3) B) 4(x−3) x−2\frac{4(x-3) }{x-2}x−24(x−3) C) 8(x−3) x−2\frac{8(x-3) }{x-2}x−28(x−3) D) 8(x−3) x+2\frac{8(x-3) }{x+2}x+28(x−3)
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