Các câu hỏi tương tự
Tính:
a)\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}}\cdot\sqrt{18}-\sqrt{128}}}\)
b)\(\sqrt{6+2\cdot\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
\(\sqrt{2\cdot x^2+4\cdot x+6}\) +\(\sqrt{3\cdot x^2+6\cdot x+12}\)=5-\(2\cdot x\)-\(x^2\)
\(y=\frac{1}{2+\sqrt{2}}+\frac{1}{3\cdot\sqrt{2}+2\cdot\sqrt{3}}+\frac{1}{4\cdot\sqrt{3}+3\cdot\sqrt{4}}+...+\frac{1}{100\cdot\sqrt{99}+99\cdot\sqrt{100}}\)tính y
TÍNH :Asqrt{3+sqrt{5+2sqrt{3}}}cdotsqrt{3-sqrt{5+2sqrt{3}}}Bsqrt{4+sqrt{8}}cdotsqrt{2+sqrt{2+sqrt{2}}}cdotsqrt{2-sqrt{2+sqrt{2}}}Csqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{3}}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{3}}}}Dleft[4+sqrt{15}right]left[sqrt{10}-sqrt{6}right]cdotsqrt{4-sqrt{15}}Eleft[3-sqrt{5}right]cdotsqrt{3+sqrt{5}}text{ }+left[3+sqrt{5}right]cdotsqrt{3-sqrt{5}}
Đọc tiếp
TÍNH :
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}\cdot\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
\(B=\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(C=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(D=\left[4+\sqrt{15}\right]\left[\sqrt{10}-\sqrt{6}\right]\cdot\sqrt{4-\sqrt{15}}\)
\(E=\left[3-\sqrt{5}\right]\cdot\sqrt{3+\sqrt{5}}\text{ }+\left[3+\sqrt{5}\right]\cdot\sqrt{3-\sqrt{5}}\)
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, \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
\(2+\sqrt{17-4\cdot\sqrt{9+4\cdot\sqrt{5}}}\)
tính :
A=\(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
Thực hiện phép tính:\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
1. Tính:
a. \(\text{[}\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}+\sqrt{\frac{1}{ab}}\text{]}\cdot\sqrt{ab}\)
b.\(\text{[}-\frac{am}{b}\cdot\sqrt{\frac{n}{m}}-\frac{ab}{n}\cdot\sqrt{mn}+\frac{a^2}{b^2}\cdot\sqrt{\frac{m}{n}}\text{]}\cdot\text{[}a^2b^2\cdot\sqrt{\frac{n}{m}}\text{]}\)