Question

Rút gọn:

\(E=x^{n-2}\left(x^2-1\right)-x\left(x^{n-1}-x^{n-3}\right)\)(với \(n\inℕ;x\ge3\))

Answer 1

\(E=x^{n-2}\left(x^2-1\right)-x\left(x^{n-1}-x^{n-3}\right)\)

\(\Leftrightarrow E=x^n-x^{n-2}-x^n+x^{n-2}\)

\(\Leftrightarrow E=0\)

Answer 2

E = x^{n - 2}(x² - 1) - x(x^{n - 1} - x^{n - 3})

E = xⁿ - x^{n - 1} - x(x^{n - 1} - x^{n - 3})

E = xⁿ - x^{n - 2} - xⁿ + x^{n - 2}

E = (xⁿ - xⁿ) + (-x^{n - 2} + x^{n - 2})

E = 0