\(\dfrac{5+7\sqrt{5}}{\sqrt{5}}+\dfrac{11+\sqrt{11}}{1+\sqrt{11}}\)
K
Khách
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NK
2
26 tháng 8 2021
`6/(sqrt11+sqrt5)-(11+sqrt11)/(sqrt11+1)+1/(2sqrt5)`
`=(6(sqrt11-sqrt5))/(11-5)-(sqrt11(sqrt11+1))/(sqrt11+1)+sqrt5/10`
`=sqrt11-sqrt5-sqrt11+sqrt5/10`
`=sqrt5/10-sqrt5=(-9sqrt5)/10`
26 tháng 8 2021
\(\dfrac{6}{\sqrt{11}+\sqrt{5}}-\dfrac{11+\sqrt{11}}{\sqrt{11}+1}+\dfrac{1}{2\sqrt{5}}\)
\(=\sqrt{11}-\sqrt{5}-\sqrt{11}+\dfrac{1}{10}\sqrt{5}\)
\(=-\dfrac{9}{10}\sqrt{5}\)
14 tháng 10 2022
\(VT=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
L
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1
\(=\sqrt{\dfrac{\left(3\sqrt{5}-1\right)\left(2\sqrt{5}-3\right)}{\left(2\sqrt{5}+3\right)\left(2\sqrt{5}-3\right)}}-\sqrt{\dfrac{\left(11+\sqrt{5}\right)\left(7+2\sqrt{5}\right)}{\left(7+2\sqrt{5}\right)\left(7-2\sqrt{5}\right)}}\)
\(=\sqrt{\dfrac{33-11\sqrt{5}}{11}}-\sqrt{\dfrac{87+29\sqrt{3}}{29}}\)
\(=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)
26 tháng 9 2018
\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)
26 tháng 8 2021
Bài 1:
\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}.4\sqrt{3}-\sqrt{3}+5.\dfrac{2\sqrt{3}}{3}=2\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{3\sqrt{3}+10\sqrt{3}}{3}=\dfrac{13\sqrt{3}}{3}\)
\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}}=\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}=-\sqrt{5}\)
\(F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}=\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}+\sqrt{\left(\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}\right)^2}-\sqrt{2}=\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}+\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)
26 tháng 8 2021
Bài 2:
Ta có: G-1
\(=\dfrac{\sqrt{x}-x+\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\dfrac{-\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le0\forall x\) thỏa mãn ĐKXĐ
hay \(G\le1\)
AH
Akai Haruma
Giáo viên
27 tháng 10 2018
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
AH
Akai Haruma
Giáo viên
27 tháng 10 2018
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
8 tháng 10 2022
a: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}\)
\(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2\left(x+2\sqrt{xy}+y\right)}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
24 tháng 8 2021
a: \(\dfrac{1}{3}\cdot\sqrt{18}-\sqrt{192}-\dfrac{\sqrt{33}}{\sqrt{11}}+3\cdot\sqrt{5\dfrac{1}{3}}\)
\(=\dfrac{1}{3}\cdot3\sqrt{2}-8\sqrt{3}-\sqrt{3}+3\cdot\dfrac{4}{\sqrt{3}}\)
\(=\sqrt{2}-7\sqrt{3}+4\sqrt{3}\)
\(=\sqrt{2}+3\sqrt{3}\)
b: Ta có: \(\sqrt{\left(2\sqrt{3}-5\right)^2}-2\cdot\sqrt{7+4\sqrt{3}}\)
\(=5-2\sqrt{3}-2\cdot\left(2+\sqrt{3}\right)\)
\(=5-2\sqrt{3}-4-2\sqrt{3}\)
\(=-4\sqrt{3}+1\)
\(\dfrac{5+7\sqrt{5}}{\sqrt{5}}+\dfrac{11+\sqrt{11}}{1+\sqrt{11}}=\dfrac{\sqrt{5}\left(\sqrt{5}+7\right)}{\sqrt{5}}+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{1+\sqrt{11}}=\sqrt{5}+7+\sqrt{11}\)
ừm mik hiểu đc r nhưng bn trình bày rõ hơn cho mik đc k?