Question

Phân tích đa thức thành nhân tử:

a) \(x-9+y-2\sqrt{xy}\)

b) \(x-5\sqrt{x}+6\)

c) \(x-2\sqrt{x}-3\)

d) \(\sqrt{x}-x^2\)

Answer 1

a) ta có : \(x-9+y-2\sqrt{xy}=\left(x-2\sqrt{xy}+y\right)-9\)

\(=\left(\sqrt{x}-\sqrt{y}\right)^2-3^2=\left(\sqrt{x}-\sqrt{y}+3\right)\left(\sqrt{x}-\sqrt{y}-3\right)\)

b) ta có : \(x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6\)

\(=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)

c) ta có : \(x-2\sqrt{x}-3=x+\sqrt{x}-3\sqrt{x}-3\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)

d) ta có : \(\sqrt{x}-x^2=\sqrt{x}\left(1-\sqrt{x}^3\right)=\sqrt{x}\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)\)

Answer 2

a) \(x-9+y-2\sqrt{xy}\)

\(=\left(x-2\sqrt{xy}+y\right)-9\)

\(=\left(\sqrt{x}-\sqrt{y}\right)^2-3^2\)

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

b) \(x-5\sqrt{x}+6\)

\(=\left(\sqrt{x}^2-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)\)

\(=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

c) \(x-2\sqrt{x}-3\)

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

\(=\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)\)

\(=\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\)

d) \(\sqrt{x}-x^2\)

\(=\sqrt{x}\left(1-\sqrt{x}^3\right)\)

\(=\sqrt{x}\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)\)