\(x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Rightarrow x^3=5+2\sqrt{13}+5-2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}.x\)

          \(=10+3x\sqrt[3]{25-52}\)

          \(=10+3x\sqrt[3]{-27}\)

           \(=10-9x\)

\(\Rightarrow x^3+9x-10=0\)

\(\Leftrightarrow x^3-x+10x-10=0\)

\(\Leftrightarrow x\left(x^2-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+10\right)=0\)

Vì \(x^2+x+10=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}>0\forall x\)

=> x - 1 = 0

=> x = 1

Thay vào A = 1²⁰¹⁵ - 1²⁰¹⁶ = 0

Vậy A = 0

Giải từ từ lần lượt các biểu thức trong dấu căn nhé:

\(\sqrt{13+\sqrt{48}}=\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}=\sqrt{\left(2\sqrt{3}+1\right)^2}=2\sqrt{3}+1\)

\(\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

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

\(B=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}\)

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

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

\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+12}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+\left(2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-1-2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{1-2\sqrt{3}+\sqrt{3}^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(1-\sqrt{3}\right)^2}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2+\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)}{6-2}\)

\(\frac{\sqrt{2+\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)}{2}\)

^(*-*)^ Đầu hàng

\(=\frac{\left(\sqrt{13}+\sqrt{11}\right)^2+\left(\sqrt{13}-\sqrt{11}\right)^2}{\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)}\)

\(=\frac{13+2\sqrt{143}+11+13-2\sqrt{143}+11}{13-11}\)

\(=\frac{48}{2}=24\)

ĐKXĐ: x>=0

Để A là số nguyên thì \(\sqrt{x}+13⋮\sqrt{x}+5\)

=>\(\sqrt{x}+5+8⋮\sqrt{x}+5\)

=>\(\sqrt{x}+5\inƯ\left(8\right)\)

mà \(\sqrt{x}+5>=5\)

nên \(\sqrt{x}+5=8\)

=>x=9

ĐK: \(x\ge0\) 

Để \(\dfrac{\sqrt{x}+13}{\sqrt{x}+5}\) có giá trị nguyên 

Mà:  \(\dfrac{\sqrt{x}+13}{\sqrt{x}+5}=\dfrac{\sqrt{x}+5+8}{\sqrt{x}+5}\)

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+5}+\dfrac{8}{\sqrt{x}+5}=1+\dfrac{8}{\sqrt{x}+5}\)

Vậy:  \(8\) ⋮ \(\sqrt{x}+5\)

\(\Rightarrow\sqrt{x}+5\inƯ\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

Mà: \(\sqrt{x}+5\ge5\)

\(\Rightarrow\sqrt{x}+5\in\left\{8\right\}\)

\(\Rightarrow x=9\left(tm\right)\)

\(\frac{\sqrt{2\sqrt{3}+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\\ =\frac{\sqrt{2\sqrt{3}+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{2\sqrt{3}+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\\ =\frac{\sqrt{2\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{2\sqrt{3}+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{3\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)

mk ko pit lm tiep dau nha 

ta có: A³=\(6\sqrt{3}+10-6\sqrt{3}+10-3\sqrt[3]{\left(6\sqrt{3}+10\right)\left(6\sqrt{3}-10\right)}.\left(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\right)\)

=\(20-3.\sqrt[3]{8}.A\)=\(20-6A\)

do đó A³=20-6A↔A³+6A-20=0↔(A²+2A+10)(A-2)=0

dễ thấy A²+2A+10>0→A=2

b) giống a)

c)giống b)