=(49-20căn6)(5+2căn6)căn(5+2căn6)/(9c...
Đầu tiên bạn nhân cả tử và mẫu với 9căn3+11căn2 , đồng thời bạn biến đổi như sau:
N=(25-2.5.2.căn6+24)(3+2.căn3.căn2+2)c... [(9căn3)^2-(11căn2)^2] =(5-2căn6)^2 . (căn3+căn2)^2 . căn(căn3+căn2)^2 . (9căn3+11căn2) /(243-242)
=(căn3-căn2)^4 . (căn3+căn2)^3 . (9căn3+11căn2) =(căn3-căn2)(9căn3+11căn2)
=27+11căn6-9căn6-22=5+2căn6.
Rút gọn:
\(A=\sqrt{11-2\sqrt{10}}+\sqrt{9-2\sqrt{4}}-\sqrt{10}-\sqrt{7}\)
\(B=\frac{\sqrt{3}+\sqrt{11+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
Toán số 9, liên quan đến dạng căn. Các bạn giúp mình nhé, xin cảm ơn rất nhiều. :)
1) \(\left(\sqrt{3-2\sqrt{\sqrt{3}-1}}+\frac{\sqrt{3}-1}{\sqrt{2}}\right)\sqrt{\sqrt{3}-1}\)
2) \(\left(\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}\right):2\sqrt{\sqrt{5}-2}\)
3) \(\frac{\sqrt{10+6\sqrt{2}}-\sqrt{10-6\sqrt{2}}}{\sqrt{5-\sqrt{7}}}-\sqrt{9+2\sqrt{20}}\)
4) \(\frac{\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}}{\sqrt{5+\sqrt{22}}}-\frac{\sqrt{6-\sqrt{24}}}{\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}}\)
5) \(\sqrt{5+2\sqrt{14\sqrt{5}-26}}-\sqrt{4\sqrt{5}-1+\sqrt{80-8\sqrt{5}}}\)
6) \(\frac{\sqrt{11+\sqrt{5}}+\sqrt{11-\sqrt{5}}}{\sqrt{11+2\sqrt{29}}}-\sqrt{3-2\sqrt{2}}\)
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) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
Rút gọn bt
\(a,A=\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(b,C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(c,\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(d,\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
Rút gọn
A=\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
B= \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
C= \(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{16}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
D=\(\sqrt{4+\sqrt{5+\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
Mn lm đc câu nào thì đc, hết càng tốt <3
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)+\(\dfrac{8}{1-\sqrt{5}}\)
\(\dfrac{5+\sqrt{7}}{9-\sqrt{23+8\sqrt{7}}}\)+\(\dfrac{5-\sqrt{7}}{2+\sqrt{16+6\sqrt{7}}}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)+\(\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
Rút gọn biểu thức:
a)\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
b)\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
c)\(5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}\)
d)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
e)\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
RÚT GỌN:\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
\(Tính\)
\(a.\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\sqrt{27-9\sqrt{5}}\)
\(b.\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{4+\sqrt{3}}{5-2\sqrt{3}}}\)
\(c.\frac{3-4\sqrt{3}}{\sqrt{6}-\sqrt{2}-\sqrt{5}}\)
\(d.\left(\sqrt{11}-\sqrt{3}\right)\left(\sqrt{13-\sqrt{6}+2\sqrt{30-\sqrt{45}}}+\sqrt{11}-\sqrt{10-\sqrt{6}}\right)\)
\(e.\frac{\sqrt{4+\sqrt{5}}+\sqrt{4-\sqrt{5}}}{\sqrt{4}+\sqrt{11}}-\frac{\sqrt{20-4\sqrt{23}}}{\sqrt{5+\sqrt{2}}-\sqrt{5-\sqrt{2}}}\)
