\(\sqrt{x}=a;a>0\Leftrightarrow A=a^3-3a^2+4a-2\)

\(\Leftrightarrow A=\left(a^3-3a^2+3a-1\right)+\left(a-1\right)\)

\(\Leftrightarrow A=\left(a-1\right)^3+\left(a-1\right)\)

\(A=\left(a-1\right)\left[\left(a-1\right)^2+1\right]\)

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

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

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

c/ \(=x-4-4\sqrt{x-4}+4=\left(\sqrt{x-4}-2\right)^2\)

d/ \(=\left(\sqrt{x}+2\right)^2\)

\(x+\sqrt{x}+2\sqrt{x}+2\)

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

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

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

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

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

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

\(b\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)\)

\(1a.\) Để : \(\sqrt{x+\dfrac{3}{x}}+\sqrt{-3x}\) xác định thì :

\(x+\dfrac{3}{x}\) ≥ 0 và \(-3x\) ≥ 0

⇔ \(\dfrac{x^2+3}{x}\) ≥ 0 và : x ≤ 0 ⇔ x > 0 và : x ≤ 0 ( Vô lý )

⇔ x ∈ ∅

b. Để : \(\sqrt{x^2+4x+5}\) xác định thì :

\(x^2+4x+5\) ≥ 0

Mà : \(x^2+4x+5=\left(x+2\right)^2+1>0\)

Vậy , ........

c. Để : \(\sqrt{2x^2+4x+5}\) xác định thì :

\(2x^2+4x+5\) ≥ 0

Mà : \(2\left(x^2+2x+1\right)+3=2\left(x+1\right)^2+3>0\)

Vậy ,.........

Bài 2. \(a.x+5\sqrt{x}+6=x+2.\dfrac{5}{2}\sqrt{x}+\dfrac{25}{4}+6-\dfrac{25}{4}=\left(\sqrt{x}+\dfrac{5}{2}\right)^2-\dfrac{1}{4}=\left(\sqrt{x}+\dfrac{5}{2}-\dfrac{1}{2}\right)\left(\sqrt{x}+\dfrac{5}{2}+\dfrac{1}{2}\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\)

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