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

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

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

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

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

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

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

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

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

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

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

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

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

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

Bài 1:

\(\frac{x-9}{\sqrt{x}+3}+\frac{2\sqrt{x}-6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

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

Bài 2:

a) Không rõ đề

b) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{3}+1\\x-3=-\sqrt{3}-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\)

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

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

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

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

mk ko pit lm tiep dau nha 

a: Ta có: \((x - \sqrt{x^2+5})(x + \sqrt{x^2+5}) = x^2 - (x^2+5) = -5\)

=>\(x + \sqrt{x^2+5} = \frac{-5}{x - \sqrt{x^2+5}}\)

Ta có: \((x - \sqrt{x^2+5})(y - \sqrt{y^2+5}) = 5\)

=>\((x-\sqrt{x^2+5})(y-\sqrt{y^2+5})\left(x+\sqrt{x^2+5}\right)=5\left(x+\sqrt{x^2+5}\right)\)

=>\(-5 (y - \sqrt{y^2+5}) = 5 (x + \sqrt{x^2+5})\)

=>\(-(y-\sqrt{y^2+5})=x+\sqrt{x^2+5}\)

=>\(\sqrt{y^2+5}-y=x+\sqrt{x^2+5}\quad(1)\)

Ta có: \((x - \sqrt{x^2+5})(y - \sqrt{y^2+5}) = 5\)

=>\((x-\sqrt{x^2+5})(y-\sqrt{y^2+5})\left(y+\sqrt{y^2+5}\right)=5\left(y+\sqrt{y^2+5}\right)\)

=>\(-5 (x - \sqrt{x^2+5}) = 5 (y + \sqrt{y^2+5})\)

=>\(\sqrt{x^2+5}-x=y+\sqrt{y^2+5}\quad(2)\)

Từ (1),(2) suy ra -x-y=x+y

=>x+y=0

=>x=-y

\(M=x^{2015}+y^{2015}=\left(-y\right)^{2015}+y^{2015}=0\)

b: Đặt \(u=\sqrt[3]{5 + 2\sqrt{13}};v=\sqrt[3]{5-2\sqrt{13}}\)

=>\(uv = \sqrt[3]{(5 + 2\sqrt{13})(5 - 2\sqrt{13})} = \sqrt[3]{5^2 - (2\sqrt{13})^2} = \sqrt[3]{25 - 52} = \sqrt[3]{-27} = -3\)

x=u+v

=>\(x^3 = (u + v)^3 = u^3 + v^3 + 3uv(u + v)\)

=>\(x^3=(5+2\sqrt{13})+(5-2\sqrt{13})+3(-3)x\)

=>\(x^3=10-9x\)

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

=>(x-1)(x^2+x+10)=0

=>x-1=0

=>x=1

\(A=x^{2015}-x^{2016}+2017\)

\(=1^{2015}-1^{2016}+2017\)

=2017