Question

a, Rút gọn: B=\(\dfrac{4x^3+8x^2-x-2}{4x^2+4x+1}\)

b, tìm x \(\in\) Z để B \(\in\) Z

Answer 1

a, \(B=\dfrac{4x^3+8x^2-x-2}{4x^2+4x+1}\)

\(=\dfrac{4x^3+2x^2+6x^2+3x-4x-2}{\left(2x+1\right)^2}\)

\(=\dfrac{2x^2\left(2x+1\right)+3x\left(2x+1\right)-2\left(2x+1\right)}{\left(2x+1\right)^2}\)

\(=\dfrac{\left(2x^2+3x-2\right)\left(2x+1\right)}{\left(2x+1\right)}\)

\(=\dfrac{2x^2+3x-2}{2x+1}\)

b, Để \(B\in Z\Leftrightarrow2x^2+3x-2⋮2x+1\)

\(\Leftrightarrow2x^2+x+2x+1-3⋮2x+1\)

\(\Leftrightarrow x\left(2x+1\right)+\left(2x+1\right)-3⋮2x+1\)

\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)-3⋮2x+1\)

\(\Leftrightarrow3⋮2x+1\)

\(\Leftrightarrow2x+1\in\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow x\in\left\{0;-1;1;-2\right\}\)

Vậy...