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

\(=\dfrac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1\)

Cau 1: 

a: \(A=\dfrac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)+2\sqrt{a}\left(\sqrt{a}-2\right)}{a-4}\)

\(=\dfrac{\left(\sqrt{a}-2\right)\left(a+4\sqrt{a}+4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}=\sqrt{a}+2\)

c: \(=\dfrac{\left|c+1\right|}{\left|c\right|-1}\)

TH1: c>0

\(C=\dfrac{c+1}{c-1}\)

TH2: c<0

\(C=\dfrac{\left|c+1\right|}{-\left(c+1\right)}=\pm1\)

\(2.p=\left(a+2016\right)-\left(a-2016\right)=2.2016\Rightarrow P=2016\)

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

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

Tham khảo nhé!

\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)

\(=\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)

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

\(=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)

Vì \(a\le6\Rightarrow\sqrt{a-2}\le2\Rightarrow\sqrt{a-2}-2\le0\Rightarrow\left|\sqrt{a-2}-2\right|=2-\sqrt{a-2}\)

Vì \(a\ge2\Rightarrow\sqrt{a-2}+2\ge2>0\)

\(\Rightarrow\text{ }\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|=\sqrt{a-2}+2+2-\sqrt{a-2}=4\)

Ta có: \(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)

\(=\sqrt{a-2}+2-\sqrt{a-2}+2\)

=4

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

b)\(7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)