Question

Phân tích đa thức thành nhân tử (với x > hoặc = 0)

a) x + \(\sqrt{x}\)

b) x - 4\(\sqrt{x}\)+ 3

Answer 1

a ) \(x+\sqrt{x}=\left(\sqrt{x}\right)^2+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right)\)

b ) \(x-4\sqrt{x}+3=\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2-1=\left(\sqrt{x}-2\right)^2-1\)

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

Answer 2

\(x+\sqrt{x}=\left(\sqrt{x}\right)^2+\sqrt{x}=\sqrt{x}.\left(\sqrt{x}+1\right)\)

\(x-4\sqrt{x}+3=\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2\right]-1^2=\left(\sqrt{x}-2\right)^2-1^2\)

\(=\left(\sqrt{x}-2-1\right)\left(\sqrt{x}-2+1\right)\)

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

Answer 3

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

b)\(x-4\sqrt{x}+3\)

\(=\left(\sqrt{x}\right)^2-2.2\sqrt{x}+2^2-1\)

\(=\left(\sqrt{x}-2\right)^2-1\)

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