mk ko thấy đề

A=(\(\sqrt{13}\).\(\sqrt{2}\)+5\(\sqrt{2}\))\(\sqrt{19-5\sqrt{13}}\)

   =(\(\sqrt{13}\)+5)\(\sqrt{2}\). \(\sqrt{19-5\sqrt{13}}\)

   =(\(\sqrt{13}\)+5) \(\sqrt{2\left(19-5\sqrt{13}\right)}\)

   = (\(\sqrt{13}\)+5) \(\sqrt{38-2.5\sqrt{13}}\)

   =(\(\sqrt{13}\)+5) \(\sqrt{5^2-2.5\sqrt{13}+13}\)

   =(\(\sqrt{13}\)+5)\(\sqrt{\left(5-\sqrt{13}\right)^2}\)

   =(\(\sqrt{13}\)+5) \(|5-\sqrt{13}|\)

   =(5+\(\sqrt{13}\))(5-\(\sqrt{13}\))

   = 25-13 = 12

\(H=cot15^o.cot35^o.cot55^o.cot75^o\)

\(=\left(cot15^o.cot75^o\right).\left(cot35^o.cot55^o\right)\)

\(=\left(cot15^o.tan15^o\right).\left(cot35^o.tan35^o\right)\)

\(=1\)

\(I=tan10^o.tan20^o.tan30^o....tan80^o\)

\(=\left(tan10^o.cot10^o\right).\left(tan20^o.cot20^o\right).\left(tan30^o.cot30^o\right).\left(tan40^o.cot40^o\right)\)

\(=1\)

\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

Tất cả đều phải tìm điều kiện

\(A=\dfrac{\sqrt{60}}{\sqrt{15}}\\=\sqrt{\dfrac{60}{15}}\\=\sqrt{4}=2\)

\(B=\sqrt{\dfrac{72}{15}}:\sqrt{\dfrac{2}{15}}\\=\sqrt{\dfrac{72}{15}}\cdot\sqrt{\dfrac{15}{2}}\\=\sqrt{\dfrac{72}{2}}=6\)

\(C=\left(\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{2}-\sqrt{3}\right)\\=\left(\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2\\=2-3=-1\)