\(\sqrt{96}-6\sqrt{\frac{2}{3}}+\frac{3}{3+\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(\sqrt{96}-6\sqrt{\frac{2}{3}}+\frac{3}{3+\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
Những câu hỏi liên quan
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, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
1)\(\sqrt{36}-6\sqrt{\frac{2}{3}}+\frac{3}{3+\sqrt{6}}-\sqrt{10-4\sqrt{6}}\) 2) \(\sqrt{7-4\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{12+6\sqrt{3}}\)
Rút gọn :Bfrac{6-6sqrt{3}}{1-sqrt{3}}+frac{3sqrt{3}+3}{sqrt{3}+1}Cfrac{3+sqrt{3}}{sqrt{3}}+frac{sqrt{6}-sqrt{3}}{1-sqrt{2}}Dfrac{sqrt{10}-sqrt{2}}{sqrt{5}-1}+frac{2-sqrt{2}}{sqrt{2}-1}Efrac{sqrt{15}-sqrt{12}}{sqrt{5}-2}+frac{1}{2-sqrt{3}}Fleft(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right)left(sqrt{6}+11right)
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Rút gọn :
\(B=\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(C=\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
\(D=\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(E=\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
\(F=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
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Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
Rút gọn biểu thức 1) frac{sqrt{5+2sqrt{6}}+sqrt{8+2sqrt{15}}}{sqrt{7+2sqrt{10}}}2) left(2+frac{3+sqrt{3}}{sqrt{3}+1}right)left(2+frac{3-sqrt{3}}{sqrt{3}-1}right):left(sqrt{5}-2right)3) left(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right).left(sqrt{6}+11right)4) frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+...+frac{1}{sqrt{99}+sqrt{100}}5) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-...-frac{1}{sqrt{98}-sqrt{99}}+frac{1}{sqrt{99}-s...
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Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
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Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
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Sao làm hổng ai bảo đú.n/g vậy :(((
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b1. Rút gọn
a)\(\frac{5\sqrt{6}+6\sqrt{5}}{\sqrt{5}+\sqrt{6}}\)
b) \(\frac{2\sqrt{7}-4\sqrt{3}}{3\sqrt{35}-6\sqrt{15}}\)
c) \(\frac{12\sqrt{10}-16\sqrt{14}}{6\sqrt{5}-8\sqrt{7}}\)
d) \(\frac{6\sqrt{6}-27}{2\sqrt{2}-3\sqrt{3}}\)
e) \(\frac{-4\sqrt{2}+3\sqrt{5}}{-2\sqrt{10}}\)
a. P (3+sqrt{2}+sqrt{6})(sqrt{6-3sqrt{3}})
b. A(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}): (sqrt{6}+11)
c. B frac{sqrt{8-2sqrt{12}}}{sqrt{3}-1}-sqrt{8}
d. C sqrt{4+sqrt{7}}-sqrt{4-sqrt{7}}-sqrt{2}
đ. Dfrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}}{3sqrt{2}+2sqrt{3}}}
e. E sqrt{8+2sqrt{10+2sqrt{5}}}+sqrt{8-2sqrt{10+2sqrt{5}}}
ê. G sqrt{4+5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}
g. Hfrac{2sqrt{4+sqrt{5+21+sqrt{80}}}}{sqrt{10}-sqrt{2}}
i. Isqrt{frac{4-sqrt{7}}{4+sqrt{7...
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a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
tính
\(\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\frac{3+6\sqrt{3}}{\sqrt{3}}-\frac{13}{\sqrt{3}+4}\)
\(3\sqrt{\frac{3}{2}}-\sqrt{6}+\sqrt{\frac{2}{3}}\)
\(\left[3-\sqrt{\left(\sqrt{3}-1\right)^2}\right]^2+\sqrt{147}\)
\(\frac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}-\frac{\sqrt{10}-\sqrt{15}}{\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}}\)
Câu 1 khai phương, rút gọn, quy đồng rồi tính được kết quả là 8+\(\sqrt{3}\)
Nói tóm lại là áp dụng các công thức biến đổi đơn giản và quy đồng là ra hết. Làm câu 2 với câu 3 trước ấy, 2 câu đấy dễ hơn.
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Giải cho câu 3, còn lại tự làm nha.
\([3-\sqrt{\left(\sqrt{3}-1\right)^2}]+\sqrt{147}\)
= \(\left[3-\left(\sqrt{3}-1\right)\right]^2+\sqrt{147}\)
= \(\left[4-\sqrt{3}\right]^2+\sqrt{147}\)
= \(16-8\sqrt{3}+9+\sqrt{147}\)
= \(25-8\sqrt{3}+\sqrt{49\cdot3}\)
= \(25-8\sqrt{3}+\sqrt{49}\cdot\sqrt{3}\)
= \(25-8\sqrt{3}+7\sqrt{3}\)
= \(25-\sqrt{3}\)
Các ý còn lại tự làm......
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Bài 3: Thực hiện phép tính
1) 2sqrt{5}-sqrt{125}-sqrt{80}+sqrt{605}
2) frac{10-2sqrt{10}}{sqrt{5}+sqrt{2}}+frac{8}{1-sqrt{5}}
3)sqrt{15-sqrt{216}}+sqrt{33-12sqrt{6}}
5)sqrt{frac{2-sqrt{3}}{2+sqrt{3}}}+sqrt{frac{2+sqrt{3}}{2-sqrt{3}}}
6)6sqrt{frac{16}{3}}-3sqrt{frac{1}{27}}-6sqrt{frac{4}{75}}
7)2sqrt{27}-6sqrt{frac{4}{3}}+frac{3}{5}sqrt{75}
8)frac{sqrt{3-sqrt{5}}.left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}}
9)sqrt{2-sqrt{3}}left(sqrt{5}+sqrt{2}right)
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Bài 3: Thực hiện phép tính
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
2) \(\frac{10-2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
3)\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
5)\(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
6)\(6\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
7)\(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
8)\(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
9)\(\sqrt{2-\sqrt{3}}\left(\sqrt{5}+\sqrt{2}\right)\)