Tính \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}-\sqrt{10}\)
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6 tháng 7 2021
1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)
2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\dfrac{\sqrt{2}}{2}\left|\sqrt{3}-\sqrt{5}\right|+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left|\sqrt{5}-1\right|\)
\(=\dfrac{\sqrt{2}}{2}\left(\sqrt{5}-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)
3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)
\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\left|\sqrt{3}-1\right|}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)
6 tháng 7 2021
3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)\)
=-2
4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
19 tháng 9 2021
\(1,B=9-5=4\\ 2,\dfrac{\sqrt{5}+1}{3-2\sqrt{2}}-\dfrac{\sqrt{10}}{\sqrt{5}-2}+3\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(\sqrt{5}+1\right)\left(3+2\sqrt{2}\right)-\sqrt{10}\left(\sqrt{5}+2\right)+3\sqrt{2}-3\sqrt{5}\\ =3\sqrt{5}+2\sqrt{10}+3+2\sqrt{2}-5\sqrt{2}-2\sqrt{10}+3\sqrt{2}-3\sqrt{5}=3\)
\(3,\\ a,\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\left(x,y\ge0;xy\ne1\right)\\ =\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\\ =\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}+\sqrt{y}-y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{1+x+y+xy}\\ =\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+x\right)+y\left(1+x\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+y\right)\left(1+x\right)}\)
\(b,x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{1}=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)
Thay vào BT
\(=\dfrac{2\left(\sqrt{3}-1+\sqrt{y}\right)}{\left(1+y\right)\left(1+4-2\sqrt{3}\right)}=\dfrac{2\sqrt{3}-2+2\sqrt{y}}{\left(1+y\right)\left(3-2\sqrt{3}\right)}\\ =\dfrac{2\sqrt{3}-2+2\sqrt{y}}{3-2\sqrt{3}+3y-2y\sqrt{3}}\)
21 tháng 4 2017
Ta có:
\(A=\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}\)
\(\Leftrightarrow A^2=10-2\sqrt{25-17}=10-4\sqrt{2}\)
\(\Leftrightarrow A=\sqrt{10-4\sqrt{2}}\)
Ta lại có:
\(B=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow B^2=6-2\sqrt{9-5}=2\)
\(\Leftrightarrow B=\sqrt{2}\)
Thế vô biểu thức ban đầu ta được
\(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
\(=\frac{\sqrt{10-4\sqrt{2}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{2}+2-\sqrt{2}}=\frac{4}{2}=2\)
VT
30 tháng 9 2019
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=\(\frac{2\sqrt{3}}{6}\)+\(\frac{\sqrt{2}}{6}\)-\(\frac{2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{6}\)=\(\frac{2\sqrt{3}+\sqrt{2}-2\left(\sqrt{3}-\sqrt{2}\right)}{6}\)=\(\frac{3\sqrt{2}}{6}\)=\(\frac{1}{\sqrt{2}}\)
5 tháng 8 2021
a) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\dfrac{2-\sqrt{3}}{1}+\dfrac{2+\sqrt{3}}{1}\)
=4
ND
0
BN
1
3 tháng 8 2023
\(=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
=(căn 5-1)(căn 5-1)(3+căn 5)
=(6-2căn 5)(3+căn 5)
=18+6căn 5-6căn 5-10
=8
6 tháng 8 2019
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
Đáp án là 0,301823955......
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