Bài 1: Căn bậc hai
Các câu hỏi tương tự
rút gọn
a) A=\(\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
b) B=\(\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
c) C=\(\left(\sqrt{3}+\sqrt{5}\right)\times\sqrt{7-2\sqrt{10}}\)
lm nhanh giúp mk nhé
(1)frac{sqrt{6+4sqrt{2}}}{sqrt{2}} (2)frac{sqrt{3-sqrt{5}}}{sqrt{0.5}} (3)left(sqrt{2}-1right)^2 (4)left(3-2sqrt{2}right).left(3+2sqrt{2}right) (5)sqrt{left(2-sqrt{3}right)}^2-sqrt{left(1-sqrt{3}right)}^2 (6)sqrt{left(sqrt{2}-sqrt{3}right)^2}-sqrt{left(sqrt{2}+sqrt{3}right)}^2 (7)frac{1}{sqrt{3}-1}-frac{1}{sqrt{3}+1} (8)sqrt{3-2sqrt{2}} (9)sqrt{4+2sqrt{3}}-sqrt{4-2sqrt{3}} (10)sqrt{2020+2sqrt{2019}}-sqrt{2020-2sqrt{2019}} (11)sqrt{7+2sqrt{12}} Các bạn giúp mình với ,Mình xin cảm ơn trước
Đọc tiếp
(1)\(\frac{\sqrt{6+4\sqrt{2}}}{\sqrt{2}}\) (2)\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{0.5}}\) (3)\(\left(\sqrt{2}-1\right)^2\) (4)\(\left(3-2\sqrt{2}\right).\left(3+2\sqrt{2}\right)\) (5)\(\sqrt{\left(2-\sqrt{3}\right)}^2-\sqrt{\left(1-\sqrt{3}\right)}^2\) (6)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)}^2\) (7)\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\) (8)\(\sqrt{3-2\sqrt{2}}\) (9)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (10)\(\sqrt{2020+2\sqrt{2019}}-\sqrt{2020-2\sqrt{2019}}\) (11)\(\sqrt{7+2\sqrt{12}}\) Các bạn giúp mình với ,Mình xin cảm ơn trước
thực hiện phép tính
A=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
B=\(\left(5+2\sqrt{6}\right)\cdot\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}\)
rÚT GỌN: Gsqrt{3+sqrt{5}}+sqrt{7-3sqrt{6}}-sqrt{2}
chững minh : a) 2sqrt{2}left(sqrt{3}-2right)+left(1+2sqrt{2}right)^2-2sqrt[]{6}9
b)left(3+sqrt{5}right)left(sqrt{10}-sqrt{2}right)sqrt{3-sqrt{5}}8
c)sqrt{dfrac{4}{left(2-sqrt{5}right)^2}}-sqrt{dfrac{4}{left(2+sqrt{5}right)^2}}8
giúp mk với tối mai mk nạp rồi
Đọc tiếp
rÚT GỌN: G=\(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{6}}-\sqrt{2}\)
chững minh : a) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt[]{6}=9\)
b)\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
c)\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
giúp mk với tối mai mk nạp rồi
a/ dfrac{sqrt{3-sqrt{5}}left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}}
b/ sqrt{8sqrt{3}}-2sqrt{25sqrt{12}}+4sqrt{sqrt{192}}
c/ sqrt{2-sqrt{3}}left(sqrt{5}+sqrt{2}right)
d/ sqrt{3-sqrt{5}}+sqrt{3+sqrt{5}}
e/ dfrac{left(sqrt{5}+2right)^2-8sqrt{5}}{2sqrt{5}-4}
Làm ơn, giúp mik với. Mik đang cần gấp lắm!
Đọc tiếp
a/ \(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
b/ \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
c/ \(\sqrt{2-\sqrt{3}}\left(\sqrt{5}+\sqrt{2}\right)\)
d/ \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
e/ \(\dfrac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
Làm ơn, giúp mik với. Mik đang cần gấp lắm!
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
B=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Rút gọn \(H=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)