Question

Phân tích đa thức thành nhân tử:

a) \(\sqrt{a^2-b^2}\) - \(\sqrt{a^3+b^3}\)

b) x - \(3\sqrt{x}\) - 18

c) \(x\sqrt{x}\) + 4x - \(12\sqrt{x}\) - 27

Answer 1

\(x\sqrt{x}+4x-12\sqrt{x}-27\)

\(=\left(x\sqrt{x}-27\right)+\left(4x-12\sqrt{x}\right)\)

\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)+4\sqrt{x}\left(\sqrt{x}-3\right)\)

\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9+4\sqrt{x}\right)\)

\(=\left(\sqrt{x}-3\right)\left(x+7\sqrt{x}+9\right)\)

Answer 2

a, \(\sqrt{a^2-b^2}-\sqrt{a^3+b^3}\)

\(=\sqrt{\left(a+b\right)\left(a-b\right)}-\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}\)

\(=\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a^2-ab+b^2}\right)\)

Answer 3

\(x-3\sqrt{x}-18\)

\(=x+3\sqrt{x}-6\sqrt{x}-18\)

\(=\sqrt{x}\left(\sqrt{x}+3\right)-6\left(\sqrt{x}+3\right)\)

\(=\left(\sqrt{x}+3\right)\left(\sqrt{x}-6\right)\)