Kết quả của phép chia (x^2 - 2x + 1) : (1 - x) là:
A. x-1 B.1-x C. -2 D.2
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\(\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(=\left(2x\right)^3+1\)
\(=8x^3+1\)
⇒ Chọn C
\(\dfrac{f\left(x\right)}{2x+1}=\dfrac{\left(2x+1\right)\left(x^2-x+1\right)}{2x+1}=x^2-x+1\)
\(\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{x^2-1}\)
\(=\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{2\left(x-1\right)-x-1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{2x-2-x-1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{x-3}{\left(x+1\right)^2\left(x-1\right)}\)
⇒Chọn B
\(\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{x^2-1}\\ =\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2.\left(x-1\right)-\left(x+1\right)}{\left(x+1\right)^2.\left(x-1\right)}\\ =\dfrac{2x-2-x-1}{\left(x+1\right)^2.\left(x-1\right)}\\ =\dfrac{x-3}{\left(x+1\right)^2\left(x-1\right)}\\ =>B\)
B