Question

\(\frac{4y^3-y}{5-10y}=\frac{-2y^2-y}{5}\)với y \(\ne\)\(\frac{1}{2}\)

Answer 1

Ta có: \(VT=\frac{4y^3-y}{5-10y}\)

\(=\frac{y\left(4y^2-1\right)}{5\left(1-2y\right)}\)

\(=\frac{y\left(2y-1\right)\left(2y+1\right)}{5\left(1-2y\right)}\)

\(=\frac{-y\left(1-2y\right)\left(2y+1\right)}{5\left(1-2y\right)}\)

\(=\frac{-y\left(2y+1\right)}{5}\)

\(=\frac{-2y^2-y}{5}=VP\)(đpcm)

Answer 2

chứng minh hai phân đa thức này bằng nhau nha