Bài 8: Rút gọn biểu thức chứa căn bậc hai
Các câu hỏi tương tự
\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
Tính:
\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6-3\sqrt{8}}\)
Tính:
\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{8}}-2\sqrt{6+3\sqrt{3}}\)
Rút Gọn Biểu Thức
B=\(\sqrt{13+3\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
C=\(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
D=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{18-\sqrt{128}}}}}\)
1 .
a)Afrac{2}{2sqrt[3]{2}+2+sqrt[3]{4}}
b)Bfrac{6}{2sqrt[3]{2}-2+sqrt[3]{4}}
c)Cfrac{2}{sqrt[3]{4}+sqrt[3]{2}+2}
2 .
a)Asqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
b)Bfrac{left(5+2sqrt{6}right).left(49-20sqrt{6}right).sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}
c)Csqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7}+4sqrt{3}}}}
d)D(left(sqrt{3}-1right).sqrt{6+2sqrt{2}.sqrt{3-sqrt{sqrt{2}+sqrt{12}+sqrt{18-sqrt{128}}}}}
Đọc tiếp
1 .
a)\(A=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
b)\(B=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\)
c)C=\(\frac{2}{\sqrt[3]{4}+\sqrt[3]{2}+2}\)
2 .
a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b)\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
c)C=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7}+4\sqrt{3}}}}\)
d)D=(\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
Rút gọn các biểu thức:
a_\(C=\frac{\sqrt{6+2\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}-\sqrt{6-2(\sqrt{6}-\sqrt{3}+\sqrt{2})}}{\sqrt{2}}\)
b_\(D=\frac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}\)
Rút gọn
a) \(\dfrac{30}{\sqrt{6}+1}+\dfrac{2}{\sqrt{6}-2}-\dfrac{6}{3-\sqrt{6}}\)
b) \(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{2}\)
c) \(\dfrac{\sqrt{21}-\sqrt{3}}{1-\sqrt{7}}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
Tính
\(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(\dfrac{1+\dfrac{\sqrt{3}}{2}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\dfrac{\sqrt{3}}{2}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}\)