\(=\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
=-2
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Các câu hỏi tương tự
a) 11+6\(\sqrt{2}\) = \(\left(3+\sqrt{2}\right)^2\)
b) 8-2\(\sqrt{7}\)=\(\left(\sqrt{7}-1\right)^2\)
c)\(\sqrt{11+6\sqrt{2}}=\sqrt{11-6\sqrt{2}}=6\)
d) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=-2\)
3 tháng 10 2021 lúc 13:53
a,\(\sqrt{8+2\sqrt{15}}\) -\(\sqrt{6+2\sqrt{15}}\)
b, \(\sqrt{17-2\sqrt{72}}-\sqrt{19+2\sqrt{18}}\)
c, \(\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}\)
d, \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)
e, \(\sqrt{10-2\sqrt{21}}-\sqrt{9-2\sqrt{14}}\)
dfrac{sqrt{3-sqrt{5}}left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}}dfrac{10+2sqrt{10}}{sqrt{5}+sqrt{2}}+dfrac{8}{1-sqrt{5}}dfrac{5+sqrt{7}}{9-sqrt{23+8sqrt{7}}}+dfrac{5-sqrt{7}}{2+sqrt{16+6sqrt{7}}}dfrac{1}{sqrt{2}+sqrt{2+sqrt{3}}}+dfrac{1}{sqrt{2}-sqrt{2+sqrt{3}}}
Đọc tiếp
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)+\(\dfrac{8}{1-\sqrt{5}}\)
\(\dfrac{5+\sqrt{7}}{9-\sqrt{23+8\sqrt{7}}}\)+\(\dfrac{5-\sqrt{7}}{2+\sqrt{16+6\sqrt{7}}}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)+\(\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
Tính
1.\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
2. \(\left(\sqrt{8}-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{18}-\sqrt{8}+\sqrt{5}\right)\)
Thực hiện phép tính:
a, M = \(\sqrt{9+4\sqrt{5}}\) - \(\sqrt{9-4\sqrt{5}}\)
b, N = \(\sqrt{8-2\sqrt{7}}\) - \(\sqrt{8+2\sqrt{7}}\)
bài 1 thực hiện pt
a)\(\sqrt{75}-2\sqrt{27}+\sqrt{48}\)
b)\(\dfrac{\left(12\sqrt{50}-8\sqrt{200}+\dfrac{7}{3}\sqrt{450}\right)}{\sqrt{10}}\)
c)\(\sqrt{8+2\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)giải hộ mik
Thực hiện phép tính:a)left(frac{2sqrt{3}-sqrt{6}}{sqrt{8}-2}-frac{sqrt{216}}{3}right).frac{1}{sqrt{6}}b) left(frac{sqrt{14}-sqrt{7}}{1-sqrt{2}}+frac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right):frac{1}{sqrt{7}-sqrt{5}}c) frac{sqrt{5-2sqrt{6}}+sqrt{8-2sqrt{15}}}{sqrt{7+2sqrt{10}}}d) left(sqrt{28}-2sqrt{14}+sqrt{7}right).sqrt{7}+7sqrt{8}
Đọc tiếp
Thực hiện phép tính:
a)\(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
b) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
d) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
Rút gọn các biểu thức sau:a. dfrac{8}{left(sqrt{5}+sqrt{3}right)^2} - dfrac{8}{left(sqrt{5}-sqrt{3}right)^2}b.dfrac{1}{4-3sqrt{2}} - dfrac{1}{4+3sqrt{2}}c.left(dfrac{sqrt{7}+3}{sqrt{7}-3}-dfrac{sqrt{7}-3}{sqrt{7}+3}right): sqrt{28}d.dfrac{3}{sqrt{6}-sqrt{3}}+dfrac{4}{sqrt{7}+sqrt{3}}
Đọc tiếp
Rút gọn các biểu thức sau:
a. \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}\) - \(\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
b.\(\dfrac{1}{4-3\sqrt{2}}\) - \(\dfrac{1}{4+3\sqrt{2}}\)
c.\(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right)\): \(\sqrt{28}\)
d.\(\dfrac{3}{\sqrt{6}-\sqrt{3}}\)+\(\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)