\(\left(\sqrt{2}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{6}+1\right)\cdot\left(5-2\sqrt{2}-\sqrt{3}\right)\)
Giúp mình giải bài trên với ạ. Cần gấp. Đầu bài yêu cầu: Tính
Thực hiện các phép tính sau:
a, \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{\sqrt{3}+2}\)
b, \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
c, \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
d, \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
Tính:
a)\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
b)\(\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{15}-4\right)\sqrt{4+\sqrt{15}}\)
Giúp mình với nha
TÍNH :
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}\cdot\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
\(B=\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(C=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(D=\left[4+\sqrt{15}\right]\left[\sqrt{10}-\sqrt{6}\right]\cdot\sqrt{4-\sqrt{15}}\)
\(E=\left[3-\sqrt{5}\right]\cdot\sqrt{3+\sqrt{5}}\text{ }+\left[3+\sqrt{5}\right]\cdot\sqrt{3-\sqrt{5}}\)
\(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
a) \(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
b) \(B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}+\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
Giúp mình với dang cần gấp
Rút gọn \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
Thực hiện phép tính:
\(a,\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(b,\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\cdot\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(c,\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(d,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(e,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(f,\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(g,\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
\(h,\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}\)