\(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)rút gọn bt
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1
KT
12 tháng 8 2018
\(A=\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\frac{6}{\sqrt{3}}\)
\(=6-3\sqrt{3}+4+\sqrt{3}+\frac{6}{\sqrt{3}}\)
\(=10-2\sqrt{3}+\frac{6}{\sqrt{3}}\)
\(=\frac{10\sqrt{3}-6+6\sqrt{3}}{\sqrt{3}}\)
\(=\frac{16\sqrt{3}-6}{\sqrt{3}}\)
26 tháng 8 2020
Ta có: \(\sqrt{2+\sqrt{3}}=\frac{1}{\sqrt{2}}.\sqrt{4+2\sqrt{3}}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}\)
=> \(A=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}}=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}}=\frac{\sqrt{3}+1}{2}\)
VA
0
4 tháng 10 2019
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{4-6}\right)-\frac{1}{\sqrt{2}}.\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}.\left(2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{2\sqrt{6}-6}{\sqrt{2}+1}-\frac{1}{\sqrt{2}}\)
17 tháng 5 2019
P=\(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\left(\frac{\sqrt{x}-3}{2\sqrt{x}-x}\right)=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{4-x}\right).\frac{2\sqrt{x}-x}{\sqrt{x}-3}=\left[\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right].\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{\left(4x+8\sqrt{x}\right).\sqrt{x}.\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\frac{4x\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\frac{4x}{\sqrt{x}-3}\)
12 tháng 10 2018
\(\frac{\sqrt{2-\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right).\)
\(=\frac{2\sqrt{2-\sqrt{3}}}{4}:\left(\frac{2\sqrt{2+\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{2\sqrt{2+\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{4}:\left(\frac{\sqrt{4+2\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{4+2\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{4}:\left[\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4\sqrt{3}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\left[\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{4\sqrt{6}}-\frac{2.4}{4\sqrt{6}}+\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4\sqrt{6}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\frac{\sqrt{18}+\sqrt{6}-8+\sqrt{6}+\sqrt{2}}{4\sqrt{6}}\)
\(=\frac{\sqrt{3}-1}{4}.\frac{4\sqrt{6}}{\sqrt{2}\left(\sqrt{9}+2\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{2}\left(\sqrt{3}+1\right)^2}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)^2}\)............
30 tháng 6 2017
=\(\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
=\(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{2}}:\left(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{6}}\right)\)
=\(\frac{\sqrt{3}+1}{2\sqrt{2}}:\left(\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}\right)\)
=\(\frac{\sqrt{3}+1}{2\sqrt{2}}:\frac{\sqrt{3}.\left(\sqrt{3}+1\right)-2.2+\sqrt{3}+1}{2\sqrt{6}}\)
=\(\frac{\sqrt{3}+1}{2\sqrt{2}.}.\frac{2\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+1}{2}\)
VD
20 tháng 4 2019
bạn nhân cả tử và mẫu với căn 2 thì mẫu có hằng đẳng thức rồi trục căn thức
Ta có :
\(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=\sqrt{1}=1\)