Chứng minh các đẳng thức sau:
a)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=2\sqrt{5}\)
b)\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=4\sqrt{2}\)
c)\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}=0\)
Những câu hỏi liên quan
Rút gọn các biểu thức sau:
a \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
b \(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
c \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
d \(\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
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rút gọn biểu thức :
A= \(\dfrac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\).
B= \(\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\).
C= \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\).
Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
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Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+2\sqrt{5}}{3\sqrt{5}-1}\)
=1
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B 4. Rút gọn các biểu thức sau:
a)\(\sqrt{6-2\sqrt{5}}\) b) \(\sqrt{3-2\sqrt{2}}\)
c)\(\sqrt{4+2\sqrt{3}}\) d)\(\sqrt{14+2\sqrt{13}}\)
a: \(=\sqrt{5}-1\)
b: \(=\sqrt{2}-1\)
c: \(=\sqrt{3}+1\)
d: \(=\sqrt{13}+1\)
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Rút gọn biểu thức:
a) dfrac{sqrt{9-2sqrt{6}}-sqrt{6}}{sqrt{3}} b)dfrac{5+sqrt{5}}{5-sqrt{5}}+dfrac{5-sqrt{5}}{5+sqrt{5}}
c) sqrt{13+30sqrt{2+sqrt{9+4sqrt{2}}}} d) sqrt{8+2sqrt{10+2sqrt{5}}}+sqrt{8-2sqrt{10+2sqrt{5}}}
e) dfrac{2sqrt{3+sqrt{5-sqrt{13+sqrt{48}}}}}{sqrt{6}-sqrt{2}} f) sqrt{9-sqrt{5sqrt{3}+5sqrt{8+10sqrt{7-4sqrt{3}}}}}
Đọc tiếp
Rút gọn biểu thức:
a) \(\dfrac{\sqrt{9-2\sqrt{6}}-\sqrt{6}}{\sqrt{3}}\) b)\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
c) \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\) d) \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
e) \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\) f) \(\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\sqrt{7-4\sqrt{3}}}}}\)
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
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thực hiện phép tínha, dfrac{sqrt{3-sqrt{5}}.left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}}b, sqrt{8sqrt{3}}-2sqrt{25sqrt{12}}+4sqrt{sqrt{192}}c, sqrt{2-sqrt{3}}.left(sqrt{5}+sqrt{2}right)d, sqrt{3-sqrt{5}}+sqrt{3+sqrt{5}}e, sqrt{4+sqrt{10+2sqrt{5}}}+sqrt{4-sqrt{10+2sqrt{5}}}f, left(5+2sqrt{6}right)left(49-20sqrt{6}right)sqrt{5-2sqrt{6}}g, dfrac{1}{sqrt{2}+sqrt{2+sqrt{3}}}+dfrac{1}{sqrt{2}-sqrt{2-sqrt{3}}}h, dfrac{6+4sqrt{2}}{sqrt{2}+sqrt{6+4sqrt{2}}}+dfrac{6-4sqrt{2}}{sqrt{2}+sqrt{6+4sqrt{2}}}i, dfrac{...
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thực hiện phép tính
a, \(\dfrac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
b, \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
c, \(\sqrt{2-\sqrt{3}}.\left(\sqrt{5}+\sqrt{2}\right)\)
d, \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
e, \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
f, \(\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}\)
g, \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
h, \(\dfrac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}\)
i, \(\dfrac{\left(\sqrt{5+2}\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
k, \(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}}\)
l, \(\dfrac{4}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)
m, \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
n, \(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3+1}}}+\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3+1}}}\)
rút gọn các biểu thức sau:
a \(\sqrt[3]{8\sqrt{5}-16}.\sqrt[3]{8\sqrt{5}+16}\)
b \(\sqrt[3]{7-5\sqrt{2}}-\sqrt[6]{8}\)
c \(\sqrt[3]{4}.\sqrt[3]{1-\sqrt{3}}.\sqrt[6]{4+2\sqrt{3}}\)
d \(\dfrac{2}{\sqrt[3]{3}-1}-\dfrac{4}{\sqrt[3]{9}-\sqrt[3]{3}+1}\)
`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
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`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
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1. sqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}+sqrt{8}
2. dfrac{sqrt{3-2sqrt{3}}}{sqrt{17-12sqrt{2}}}-dfrac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
3.sqrt{7+2sqrt{6}}-sqrt{left(sqrt{6-1}right)^2}
4sqrt{5-2sqrt{6}}-sqrt{5+sqrt{24}}
5.sqrt{4sqrt{5+sqrt{3+5sqrt{48-10sqrt{7+4sqrt{3}}}}}}
6.sqrt{6+2sqrt{5-sqrt{13+sqrt{48}}}}
Đọc tiếp
1. \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\)
2. \(\dfrac{\sqrt{3-2\sqrt{3}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
3.\(\sqrt{7+2\sqrt{6}}-\sqrt{\left(\sqrt{6-1}\right)^2}\)
4\(\sqrt{5-2\sqrt{6}}-\sqrt{5+\sqrt{24}}\)
5.\(\sqrt{4\sqrt{5+\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
6.\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
1. sqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}+sqrt{8}
2. dfrac{sqrt{3-2sqrt{3}}}{sqrt{17-12sqrt{2}}}-dfrac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
3.sqrt{7+2sqrt{6}}-sqrt{left(sqrt{6}-1right)^2}
4sqrt{5-2sqrt{6}}-sqrt{5+sqrt{24}}
5.sqrt{4sqrt{5+sqrt{3+5sqrt{48-10sqrt{7+4sqrt{3}}}}}}
6.sqrt{6+2sqrt{5-sqrt{13+sqrt{48}}}}
Đọc tiếp
1. \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\)
2. \(\dfrac{\sqrt{3-2\sqrt{3}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
3.\(\sqrt{7+2\sqrt{6}}-\sqrt{\left(\sqrt{6}-1\right)^2}\)
4\(\sqrt{5-2\sqrt{6}}-\sqrt{5+\sqrt{24}}\)
5.\(\sqrt{4\sqrt{5+\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
6.\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
Bài 1: Tính
1, Aleft(1-frac{5+sqrt{5}}{1+sqrt{5}}right).left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
2, Bleft(frac{3sqrt{125}}{15}-frac{10-4sqrt{6}}{sqrt{5}-2}right).frac{1}{sqrt{5}}
3, Cleft(frac{sqrt{1000}}{100}-frac{5sqrt{2}-2sqrt{5}}{2sqrt{5}-8}right).frac{sqrt{10}}{10}
4, Dfrac{1}{sqrt{49+20sqrt{6}}}-frac{1}{sqrt{49-20sqrt{6}}}+frac{1}{sqrt{7-4sqrt{3}}}
5, Efrac{1}{sqrt{4-2sqrt{3}}}-frac{1}{sqrt{7-sqrt{48}}}+frac{3}{sqrt{14-6sqrt{5}}}
6, Ffrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}...
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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, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)