Tính :
a, \(B=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
b, \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, \(C=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}-\frac{\sqrt{5-2\sqrt{6}}}{3}\)
Tính :
a, \(B=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
b, \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, \(C=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}-\frac{\sqrt{5-2\sqrt{6}}}{3}\)
a)=\(\sqrt{3-\sqrt{5}}\).\(\sqrt{3+\sqrt{5}}\).\(\sqrt{2}\)(\(\sqrt{5}\)-\(1\))\(\sqrt{3+\sqrt{5}}\)=2\(\sqrt{2}\) \(\sqrt{\left(\sqrt{5}-1\right)^2.\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\) .\(\sqrt{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\)\(\sqrt{8}\) =8
b)A2=8+2 căn[\(\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)\)]=8+2\(\sqrt{6-2\sqrt{5}}\)=8+2(\(\sqrt{5}\)-1)=6+2\(\sqrt{5}\)=(\(\sqrt{5}+1\))2 =>A=\(\sqrt{5}\)+1
c)C=\(\frac{2\sqrt{3}}{6}\)+\(\frac{\sqrt{2}}{6}\)-\(\frac{2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{6}\)=\(\frac{2\sqrt{3}+\sqrt{2}-2\left(\sqrt{3}-\sqrt{2}\right)}{6}\)=\(\frac{3\sqrt{2}}{6}\)=\(\frac{1}{\sqrt{2}}\)
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Tính :
a)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
b)\(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
c) \(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}}+1}-\sqrt{3-2\sqrt{2}}\)
Thực hiện phép tính
a, \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
b. \(\frac{\sqrt{3-\sqrt{5}.\left(3+\sqrt{5}\right)}}{\sqrt{10}+\sqrt{2}}\)
c, \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
d, \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
a/\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=2\sqrt{5}+\frac{8}{1-\sqrt{5}}\)
\(=\frac{2\sqrt{5}-10+8}{1-\sqrt{5}}=\frac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\)
b/Đề sai
c/\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}=\sqrt{2}\left(\frac{3+\sqrt{3}+3-\sqrt{3}}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\right)=\frac{6\sqrt{2}}{6}=\sqrt{2}\)
d/ \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9+4\sqrt{5}-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)
Thực hện phép tính
a, \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
b,\(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
c, \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
d, \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)
a: \(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-\sqrt{6}}{3}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{2}\)
b: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
d: \(=-\left(\sqrt{5}+\sqrt{2}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)=-3\)
\(Tính\)
\(a.\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\sqrt{27-9\sqrt{5}}\)
\(b.\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{4+\sqrt{3}}{5-2\sqrt{3}}}\)
\(c.\frac{3-4\sqrt{3}}{\sqrt{6}-\sqrt{2}-\sqrt{5}}\)
\(d.\left(\sqrt{11}-\sqrt{3}\right)\left(\sqrt{13-\sqrt{6}+2\sqrt{30-\sqrt{45}}}+\sqrt{11}-\sqrt{10-\sqrt{6}}\right)\)
\(e.\frac{\sqrt{4+\sqrt{5}}+\sqrt{4-\sqrt{5}}}{\sqrt{4}+\sqrt{11}}-\frac{\sqrt{20-4\sqrt{23}}}{\sqrt{5+\sqrt{2}}-\sqrt{5-\sqrt{2}}}\)
1. Rút gọn \(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\)
2. Tính \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
3.Tính \(C=\frac{\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\cdot\left(3+\sqrt{5}\right)}{\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)
Bài 2:
Ta có: \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
\(=\frac{\sqrt{\sqrt{5}-1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}-\left(\sqrt{2}-1\right)\)
\(=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\frac{4}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\sqrt{2}-\sqrt{2}+1\)
=1
câu 1. đkxđ: \(x\ge\frac{1}{2}\)
\(A\sqrt{2}=\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
\(=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(=\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)
nếu \(\left|\sqrt{2x-1}-1\right|=\sqrt{2x-1}-1\) với \(\sqrt{2x-1}\ge1\Leftrightarrow x\ge1\)
thì \(A\sqrt{2}=\sqrt{2x-1}+1-\sqrt{2x-1}+1=2\)
=> A=\(\sqrt{2}\)
nếu \(\left|\sqrt{2x-1}-1\right|=1-\sqrt{2x-1}\) với \(\frac{1}{2}\le x< 1\)
thì \(A\sqrt{2}=\sqrt{2x-1}+1-1+\sqrt{2x-1}=2\sqrt{2x-1}\)
=> A= \(\sqrt{4x-2}\)
câu 3: C = \(\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)}{\left(\text{4+\sqrt{15}}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-\sqrt{15}}}\)
\(=\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}}{\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)
=\(\frac{\sqrt{9-\left(\sqrt{5}\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}}{\sqrt{16-\left(\sqrt{15}\right)^2}.\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{15}}}\)
\(=\frac{2\left(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\right)}{\sqrt{40+10\sqrt{15}}-\sqrt{24-6\sqrt{15}}}\)
\(=2.\frac{\left(\sqrt{5}+5\right)-\left(\sqrt{5}+1\right)}{\left(\sqrt{15}+5\right)-\left(\sqrt{15}+3\right)}\)
= 4
tinh
a. \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)
b.\(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)
c.\(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)
d.\(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)
e.\(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)
f.\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
\(a,\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{9-4\sqrt{5}}\)
\(b,\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
\(c,\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{2}{5+2\sqrt{7}}\)
a) \(=\left|\sqrt{5}-3\right|+\sqrt{5-2\sqrt{20}+4}\)
\(=3-\sqrt{5}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=3-\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=3-\sqrt{5}+\sqrt{5}-2\)
\(=1\)
b)\(=\left(\sqrt{5}+1\right)\cdot\sqrt{2}\cdot\left(6-2\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2\)
\(=16\)
c)\(=\frac{\sqrt{7}-5}{2}-\frac{6\left(\sqrt{7}+2\right)}{7-4}+\frac{\left(3-\sqrt{7}\right)}{9-7}+\frac{2\left(5-2\sqrt{7}\right)}{25-28}\)
\(=\frac{\sqrt{7}-5}{2}-\frac{6\left(\sqrt{7}+2\right)}{3}+\frac{3-\sqrt{7}}{2}-\frac{2\left(5-2\sqrt{7}\right)}{3}\)
\(=\frac{\sqrt{7}-\sqrt{7}-5+3}{2}-\frac{6\sqrt{7}+12+10-4\sqrt{7}}{3}\)
\(=-1-\frac{2\sqrt{7}+22}{3}\)
\(=\frac{-3-2\sqrt{7}-22}{3}\)
\(=-\frac{25+2\sqrt{7}}{3}\)