\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

nhầm 

\(=\frac{a}{a-16}+\frac{2}{a-\sqrt{4}}-\frac{2}{a+\sqrt{4}}=\frac{a+2\left(a+\sqrt{4}\right)-2\left(a-\sqrt{4}\right)}{a-16}\)

\(=\frac{a+2a+2\sqrt{4}-2a+2\sqrt{4}}{a-16}=\frac{a+8}{a-16}\)

\(=\frac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}-2}{\sqrt{2}}\)

\(=\frac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}=\frac{0}{2}=0\)

Áp dụng hằng đẳng thức (a + b)³ = a³ + b³ + 3ab(a + b)

\(A^3=182+\sqrt{33125}+182-\sqrt{33125}+3\sqrt[3]{182+\sqrt{33125}}.\sqrt[3]{182-\sqrt{33125}}.\left(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\right)\)

\(A^3=364+3\sqrt[3]{182^2-33125}.A\)

A³ = 364 - 3A

<=> A³ + 3A - 364 = 0 

<=> A³ - 7A² + 7A² - 49A + 52A  364 = 0

<=> A². (A - 7) + 7A.(A - 7) + 52.(A - 7)= 0

<=> (A - 7).(A² + 7A + 52) = 0 

<=> A = 7 hoặc A² + 7A + 52 = 0  (*)

Giải (*) <=> (A+ 3,5)² + 39,75 = 0 (vô nghiệm)

Vậy A = 7