a) \(A=\sqrt{64}+4\sqrt{4}+2016=\sqrt{8^2}+4.\sqrt{2^2}+2016=8+4.2+2016=2032\)

b) \(B=2\sqrt{8}-3\sqrt{18}+4\sqrt{128}-5\sqrt{32}=4\sqrt{2}-9\sqrt{2}+32\sqrt{2}-20\sqrt{2}\)

\(=\sqrt{2}\left(4-9+32-20\right)=7\sqrt{2}\)

a,

\(A=\sqrt{8}^2+2.\sqrt{8}.\sqrt{2}+\sqrt{2}^2+2014\)

\(=\left(\sqrt{8}+\sqrt{2}\right)^2+2014\)

a) \(A=\left(1-\sqrt{18}+\sqrt{32}\right).\sqrt{3-2\sqrt{2}}\)

\(=\left(1-\sqrt{9.2}+\sqrt{16.2}\right).\sqrt{2-2\sqrt{2}+1}\)

\(=\left(1-\sqrt{9}.\sqrt{2}+\sqrt{16}.\sqrt{2}\right).\sqrt{\left(\sqrt{2}-1\right)^2}\)

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

\(=\left(1+\sqrt{2}\right).\left|\sqrt{2}-1\right|\)

Vì \(\sqrt{2}>1\)\(\Rightarrow\left|\sqrt{2}-1\right|>0\)

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

b) \(B=\frac{3}{6+\sqrt{35}}-\frac{3}{6-\sqrt{35}}=\frac{3\left(6-\sqrt{35}\right)}{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}-\frac{3\left(6+\sqrt{35}\right)}{\left(6-\sqrt{35}\right)\left(6+\sqrt{35}\right)}\)

\(=\frac{18-3\sqrt{35}-18-3\sqrt{35}}{36-35}=-6\sqrt{35}\)

\(A=2\sqrt{5}-\sqrt{45}+2\sqrt{20}=2\sqrt{5}-\sqrt{3^2.5}+2\sqrt{2^2.5}=2\sqrt{5}-3\sqrt{5}+4\sqrt{5}=3\sqrt{5}\)

\(B=\left(\sqrt{18}-\frac{1}{2}\cdot\sqrt{32}+12\sqrt{2}\right):\sqrt{2}=\left(3\sqrt{2}-\frac{1}{2}\cdot4\sqrt{2}+12\sqrt{2}\right):\sqrt{2}\)

\(=13\sqrt{2}:\sqrt{2}=13\)

\(C=\left(\sqrt{12}+2\sqrt{27}-3\sqrt{3}\right)\cdot\sqrt{3}=\left(2\sqrt{3}+6\sqrt{3}-3\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)

\(D=\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)

\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)

\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)

\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)

\(=\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\sqrt{5}+1\)

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

a: \(=\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+6\sqrt{2}\right)\)

\(=\sqrt{2}\cdot4\sqrt{2}=8\)

b: \(=2-\sqrt{6-2\sqrt{5}}\)

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

mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia

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