tinh y
\(y=\sqrt{5+\sqrt{3+\sqrt{5+\sqrt{3+}...}}}\)
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26 tháng 8 2021
\(A=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=x-\sqrt{xy}+y\)
\(B=\dfrac{\sqrt{x}-\sqrt{y}}{x\sqrt{x}-y\sqrt{y}}=\dfrac{1}{x+\sqrt{xy}+y}\)
\(C=\dfrac{3\sqrt{3}+x\sqrt{x}}{3-\sqrt{3x}+x}=\sqrt{x}+\sqrt{3}\)
\(D=\dfrac{x+\sqrt{5x}+5}{x\sqrt{x}-5\sqrt{5}}=\dfrac{1}{\sqrt{x}-\sqrt{5}}\)
20 tháng 9 2021
Câu a bạn xem lại đề nhé vì \(\sqrt{2-6}=\sqrt{-4}\left(VLý\right)\)
b) \(\dfrac{3\sqrt{5}+5\sqrt{3}}{\sqrt{3}+\sqrt{5}}=\dfrac{\sqrt{3.5}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}=\sqrt{15}\)
c) \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
30 tháng 6 2018
\(a.\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)}=\sqrt{x}-\sqrt{y}+\sqrt{x}+\sqrt{y}=2\sqrt{x}\)
\(b.\sqrt{\left(\sqrt{5}-1\right)\sqrt{13-\sqrt{49-2.7.2\sqrt{5}+20}}}=\sqrt{\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}}=\sqrt{\left(\sqrt{5}-1\right)\left(\sqrt{5+1}\right)}=\sqrt{5}-1\)
\(c.\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{2}.\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}=\dfrac{2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{3+2\sqrt{3}+1}}=2\left(9-5\right)=2.4=8\)
TT
1 tháng 7 2018
Câu a
\(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\sqrt{x}+\sqrt{y}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x^2y}+\sqrt{xy^2}}{\sqrt{xy}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\\ =\dfrac{2x\sqrt{y}}{\sqrt{xy}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)
6 tháng 7 2018
\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3+1+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3+1-2\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\\ =\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\\ =\dfrac{\left(2\sqrt{2}+\sqrt{6}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{\left(2\sqrt{2}-\sqrt{6}\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\dfrac{2\sqrt{6}+3\sqrt{2}-2\sqrt{2}-\sqrt{6}+2\sqrt{6}-3\sqrt{2}+2\sqrt{2}-\sqrt{6}}{2\sqrt{3}}\\ =\dfrac{4\sqrt{6}-2\sqrt{6}}{2\sqrt{3}}=\dfrac{\sqrt{6}\left(4-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}}{2\sqrt{3}}=\sqrt{2}\)
\(y=\dfrac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{6+2\sqrt{5}}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{6-2\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5+1+2\sqrt{5}}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5+1-2\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5}+1}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5}-1}\\ =\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{5}-1}\\ =\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\dfrac{\left(3\sqrt{2}-\sqrt{10}\right)\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\\ =\dfrac{3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}-3\sqrt{10}+5\sqrt{2}-3\sqrt{2}+\sqrt{10}}{4}\\ =\dfrac{4\sqrt{2}}{4}=\sqrt{2}\)
Vậy \(x=y\)
Y
8 tháng 7 2023
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
2
22 tháng 12 2022
`[\sqrt{27}-\sqrt{15}]/[3-\sqrt{5}]+4/[2+\sqrt{3}]-6/\sqrt{3}`
`=[\sqrt{3}(3-\sqrt{5})]/[3-\sqrt{5}]+[4(2-\sqrt{3})]/[4-3]-[2\sqrt{3}.\sqrt{3}]/\sqrt{3}`
`=\sqrt{3}+8-4\sqrt{3}-2\sqrt{3}`
`=8-5\sqrt{3}`
_______________________________________
`[x-y]/[\sqrt{x}+\sqrt{y}]-[x\sqrt{y}+y\sqrt{x}]/\sqrt{xy}` `ĐK: x,y > 0`
`=[(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})]/[\sqrt{x}+\sqrt{y}]-[\sqrt{xy}(\sqrt{x}+\sqrt{y})]/\sqrt{xy}`
`=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}`
`=-2\sqrt{y}`
ta có: \(y^2=5+\sqrt{3+\sqrt{5+\sqrt{3+...}}}.\)
\(\Leftrightarrow\left(y^2-5\right)^2=3+\sqrt{5+\sqrt{3+\sqrt{5+\sqrt{3+...}}}}\)
\(\Leftrightarrow y^4-10y^2+25-3=y\) vì \(y=\sqrt{5+\sqrt{3+\sqrt{5+\sqrt{3+...}}}}\)
\(\Leftrightarrow y^4-10y^2-y+22=0\)
giải pt cuối ta đc: y=...
bn tự giải nha!
ban kia lam dung roi do
k tui nha
thanks