rút gọn \(\frac{1}{2-\sqrt{5}}+\frac{2}{\sqrt{5}+\sqrt{3}}:\frac{1}{\sqrt{21-12\sqrt{3}}}\)
\(\frac{5\sqrt{3}}{\sqrt{3}-\sqrt{5}-\sqrt{3}}-\frac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}+\sqrt{3}}\)
Những câu hỏi liên quan
rút gọn các căn thức sauBfrac{sqrt{5-sqrt{3}}-sqrt{5+sqrt{3}}}{sqrt{5-sqrt{22}}}+sqrt{27+10sqrt{2}}C frac{sqrt{45+27sqrt{2}}+sqrt{45-27sqrt{2}}}{sqrt{5+3sqrt{2}}-sqrt{5-3sqrt{2}}}-frac{sqrt{3+sqrt{2}}+sqrt{3-sqrt{2}}}{sqrt{3+sqrt{2}}-sqrt{3-sqrt{2}}}Dfrac{1}{sqrt{2}-sqrt{3}}.sqrt{frac{3sqrt{2}-2sqrt{3}}{3sqrt{2}+2sqrt{3}}}A frac{1}{sqrt{3+2sqrt{2}}}+frac{1}{sqrt{5+2sqrt{6}}}+frac{1}{sqrt{7+2sqrt{12}}}+....+frac{1}{sqrt{199+2sqrt{9900}}}
Đọc tiếp
rút gọn các căn thức sau
B=\(\frac{\sqrt{5-\sqrt{3}}-\sqrt{5+\sqrt{3}}}{\sqrt{5-\sqrt{22}}}+\sqrt{27+10\sqrt{2}}\)C= \(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)D=\(\frac{1}{\sqrt{2}-\sqrt{3}}.\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)A= \(\frac{1}{\sqrt{3+2\sqrt{2}}}+\frac{1}{\sqrt{5+2\sqrt{6}}}+\frac{1}{\sqrt{7+2\sqrt{12}}}+....+\frac{1}{\sqrt{199+2\sqrt{9900}}}\)
a,\(\frac{4}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
b,\(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\cdot\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)\)rút gọn
Rút gọn biểu thức: \(P=\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}-\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}+\frac{\left(\sqrt{5}-1\right).\sqrt[3]{2+\sqrt{5}}}{\sqrt{28}-10\sqrt{3}+\sqrt{3}}\)
Giúp mk nha!
Xem kỹ lại đề nhé! loại này đề lệch một tý thôi -->Không rút được !
p/s: Tránh truongf hợp làm đến cuối mới biết đề sai.
Đúng 0
Bình luận (0)
U = \(\left(\frac{1}{2-\sqrt{5}}+\frac{2}{\sqrt{5}+\sqrt{3}}\right):\frac{1}{\sqrt{21-12\sqrt{3}}}\)
*W = \(\frac{5\sqrt{3}}{\sqrt{3-\sqrt{5}-\sqrt{3}}}-\frac{5\sqrt{3}}{\sqrt{3-\sqrt{5}+\sqrt{3}}}\)
Y = \(\frac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
Rút gọn biểu thức 1)frac{15}{3sqrt{20}}2) frac{sqrt{15}-sqrt{6}}{sqrt{2}-sqrt{5}}3) frac{3}{sqrt{5}-sqrt{2}}+frac{4}{sqrt{6}+sqrt{2}}4) sqrt{frac{3}{20}}+sqrt{frac{1}{60}}-2sqrt{frac{1}{15}}5) left(sqrt{20}-sqrt{45}+sqrt{5}right)sqrt{5}6)left(2+sqrt{5}right)^2-left(2+sqrt{5}right)^27) left(5sqrt{frac{1}{5}}+frac{1}{2}sqrt{20}-frac{5}{4}sqrt{frac{4}{5}+sqrt{5}}right):2sqrt{5}8)frac{1}{3}sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1frac{1}{3}}9) 2sqrt{3}left(2sqrt{6}-sqrt{3}+1right)10) left(5sqrt{3}+3sqrt{...
Đọc tiếp
Rút gọn biểu thức
1)\(\frac{15}{3\sqrt{20}}\)
2) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{2}-\sqrt{5}}\)
3) \(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}\)
4) \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
5) \(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right)\sqrt{5}\)
6)\(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
7) \(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)
8)\(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
9) \(2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)\)
10) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
11) \(\sqrt{\sqrt{10}+1}.\sqrt{\sqrt{10}-1}\)
12) \(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
13) \(\sqrt{\frac{3}{4}}+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{12}}\)
14) \(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}\right)\sqrt{6}\)
15 ) \(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
16) \(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
17) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
Đúng 0
Bình luận (0)
rút gọn
\(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(\frac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}-\frac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
11) frac{3}{sqrt{6}-sqrt{3}}+frac{4}{sqrt{7}+sqrt{3}}
12) frac{6}{3sqrt{2}+2sqrt{3}}
13) left(sqrt{75}-3sqrt{2}-sqrt{12}right)left(sqrt{3}+sqrt{2}right)
14)frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}+frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}
15)frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}+frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}-frac{sqrt{5}+1}{sqrt{5}-1}
16)frac{sqrt{2}}{2sqrt{3}+4sqrt{2}}
17) frac{1}{4-3sqrt{2}}-frac{1}{4+3sqrt{2}}
18)frac{6}{sqrt{2}-sqrt{3}+3}
19)frac{sqrt{3+2sqrt{2}}+sqrt{3-2sqrt{2}...
Đọc tiếp
11) \(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
12) \(\frac{6}{3\sqrt{2}+2\sqrt{3}}\)
13) \(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
14)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
15)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
16)\(\frac{\sqrt{2}}{2\sqrt{3}+4\sqrt{2}}\)
17) \(\frac{1}{4-3\sqrt{2}}-\frac{1}{4+3\sqrt{2}}\)
18)\(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
19)\(\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\)
20)\(\sqrt{24}+6\sqrt{\frac{2}{3}}+\frac{10}{\sqrt{6}-1}\)
21)\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{58}}\)
22)\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)
23)\(\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right):\left(3\sqrt{18}-2\sqrt{27}+\sqrt{45}\right)\)
24)\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
25)\(\left(\sqrt{7}-\sqrt{5}\right)^2+2\sqrt{35}\)
26)\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}+\frac{3\sqrt{45}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)
27)\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}}-1}\)
28)\(\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{3+\sqrt{3}}\)
29)\(\frac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
30)\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
31)\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
32)\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)
Rút Gọn Afrac{left(frac{1}{4}-frac{sqrt{2}}{7}+frac{3sqrt{2}}{35}right)frac{1}{25}}{left(frac{1}{10}+frac{3sqrt{2}}{35}-frac{sqrt{2}}{5}right)frac{5}{7}} Bfrac{3+sqrt{5}}{sqrt{10}+sqrt{3+sqrt{5}}}-frac{3-sqrt{5}}{sqrt{10}+sqrt{3+sqrt{5}}} Cfrac{3+sqrt{5}}{sqrt{2}+sqrt{3+sqrt{5}}}-frac{3-sqrt{5}}{sqrt{2}-sqrt{3-sqrt{5}}}
Đọc tiếp
Rút Gọn A=\(\frac{\left(\frac{1}{4}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\frac{1}{25}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{35}-\frac{\sqrt{2}}{5}\right)\frac{5}{7}}\)
B=\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
C=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Rút gọn :Bfrac{6-6sqrt{3}}{1-sqrt{3}}+frac{3sqrt{3}+3}{sqrt{3}+1}Cfrac{3+sqrt{3}}{sqrt{3}}+frac{sqrt{6}-sqrt{3}}{1-sqrt{2}}Dfrac{sqrt{10}-sqrt{2}}{sqrt{5}-1}+frac{2-sqrt{2}}{sqrt{2}-1}Efrac{sqrt{15}-sqrt{12}}{sqrt{5}-2}+frac{1}{2-sqrt{3}}Fleft(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right)left(sqrt{6}+11right)
Đọc tiếp
Rút gọn :
\(B=\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(C=\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
\(D=\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(E=\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
\(F=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
Đúng 0
Bình luận (1)