a: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
Rút gọn :
\(A=\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{5}}}}\)
\(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(D=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(E=\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
Tính
A=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
B=\(\left(3-\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\cdot\sqrt{3-\sqrt{5}}\)
C=\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{ }}3}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
E=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{5}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
tính:
P=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Q=\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
GIúp mình với
Rút gọn các biểu thức sau :
a)\(\left[\left(a-b\right)\sqrt{\frac{a+b}{a-b}}+a-b\right]\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)với a > b > 0
b)\(\frac{\sqrt{7-4\sqrt{3}}}{\sqrt{2-\sqrt{3}}}.\sqrt{2+\sqrt{3}}\)
Chứng minh rằng
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
Bài 1 : Thực hiện phép tính :
a ) \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}\)
b ) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
c ) \(\sqrt{12}+2\sqrt{75}-3\sqrt{48}-\frac{2}{7}\sqrt{147}\)
d ) \(\sqrt{\left(3+\sqrt{5}\right)^2}-\sqrt{9-4\sqrt{5}}\)
e ) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{\sqrt{5}+\sqrt{2}}{3}\)
f ) \(\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
g ) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}\)
h ) \(\left(\sqrt{56}-2\sqrt{6}-\sqrt{14}\right)\sqrt{14}+\sqrt{84}\)
k ) \(\left(\frac{1}{1-\sqrt{3}}-\frac{1}{1+\sqrt{3}}\right).\left(\sqrt{3}-1\right)\)
l ) \(\sqrt{21+8\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)
m ) \(\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
n ) \(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
Làm không nổi thì câu nào biết thì làm làm từ từ dần dần giúp nha các bạn
Rút gọn :
\(B=\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(C=\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
\(D=\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(E=\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
\(F=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
Tính
1, a = \(\sqrt[3]{45+26\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
2, x = \(\sqrt[3]{4+\sqrt{80}-\sqrt[3]{\sqrt{80}-4}}\)
3, \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
4, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
5, \(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
Rút gọn biểu thức :
\(\frac{\sqrt{7-4\sqrt{3}}}{\sqrt{2-\sqrt{3}}}.\sqrt{2+\sqrt{3}}\)
\(\left[\left(a-b\right)\sqrt{\frac{a+b}{a-b}}+a-b\right]\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)với a>b>0
Chứng minh rằng :
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
\(a,\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{9-4\sqrt{5}}\)
\(b,\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
\(c,\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{2}{5+2\sqrt{7}}\)
