(1+\(\sqrt{2}\)-\(\sqrt{3}\))(1+\(\sqrt{2}\)+3)
KM
Những câu hỏi liên quan
1) dfrac{2}{sqrt{5}-2}+dfrac{-2}{sqrt{5}+2}2) dfrac{4}{1-sqrt{3}}+dfrac{sqrt{3}-1}{sqrt{3}+1}3) dfrac{sqrt{2}-1}{sqrt{2}+1}-dfrac{3-sqrt{2}}{3+sqrt{2}}4) dfrac{6}{1-sqrt{3}}-dfrac{3sqrt{3}-3}{sqrt{3}+1}5) dfrac{sqrt{5}+sqrt{6}}{sqrt{5}-sqrt{6}}+dfrac{sqrt{6}-sqrt{5}}{sqrt{6}+sqrt{5}}
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1) \(\dfrac{2}{\sqrt{5}-2}+\dfrac{-2}{\sqrt{5}+2}\)
2) \(\dfrac{4}{1-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
3) \(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3-\sqrt{2}}{3+\sqrt{2}}\)
4) \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}-3}{\sqrt{3}+1}\)
5) \(\dfrac{\sqrt{5}+\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\dfrac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}\)
4: Ta có: \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(=-3-3\sqrt{3}-3\)
\(=-6-3\sqrt{3}\)
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Tính:1) dfrac{3}{1-sqrt{2}}+dfrac{sqrt{2}-1}{sqrt{2}+1}2) dfrac{sqrt{5}-1}{sqrt{5}+1}+dfrac{6}{1-sqrt{5}}3) dfrac{sqrt{2}+sqrt{3}}{2-sqrt{6}}+dfrac{sqrt{3}-sqrt{2}}{sqrt{6}+2}4) dfrac{3+sqrt{3}}{sqrt{3}}+dfrac{sqrt{6}-sqrt{3}}{1-sqrt{2}}5) dfrac{2-sqrt{2}}{1-sqrt{2}}+dfrac{sqrt{2}-sqrt{6}}{sqrt{3}-1}
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Tính:
1) \(\dfrac{3}{1-\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
2) \(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}\)
3) \(\dfrac{\sqrt{2}+\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}\)
4) \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
5) \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=-\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
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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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Thực hiện phép tínha) dfrac{3}{sqrt{2}}+sqrt{dfrac{1}{2}}-2sqrt{18}+sqrt{left(1-sqrt{2}right)^2}b) dfrac{1}{sqrt{3}}+dfrac{1}{3sqrt{2}}+dfrac{1}{sqrt{3}}dfrac{sqrt{3}-sqrt{2}}{2sqrt{3}}c) sqrt[3]{dfrac{3}{4}}.sqrt[3]{dfrac{9}{16}}d) dfrac{sqrt[3]{54}}{sqrt[3]{-2}}e) sqrt[3]{5sqrt{2}+7}-sqrt[3]{5sqrt{2}+7}
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Thực hiện phép tính
a) \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
b) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
c) \(\sqrt[3]{\dfrac{3}{4}}.\sqrt[3]{\dfrac{9}{16}}\)
d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\)
e) \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}+7}\)
a: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=\dfrac{3}{2}\sqrt{2}+\dfrac{1}{2}\sqrt{2}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)
\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)
b: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{18}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)
c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)
d: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=-\sqrt[3]{27}=-3\)
e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}+7}=0\)
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Tìm giá trị của biểu thức;
a,sqrt{5}left(sqrt{6}+1right):frac{sqrt{2sqrt{3}+sqrt{2}}}{sqrt{2sqrt{3}-sqrt{2}}}
b,frac{sqrt{3}}{1-sqrt{sqrt{3}+1}}+frac{sqrt{3}}{1+sqrt{sqrt{3}+1}}
c,frac{1}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{1}{sqrt{2}-sqrt{2-sqrt{3}}}
d,frac{2+sqrt{3}}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{2-sqrt{3}}{sqrt{2}-sqrt{2-sqrt{3}}}
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Tìm giá trị của biểu thức;
a,\(\sqrt{5}\left(\sqrt{6}+1\right):\frac{\sqrt{2\sqrt{3}+\sqrt{2}}}{\sqrt{2\sqrt{3}-\sqrt{2}}}\)
b,\(\frac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}+\frac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
c,\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
d,\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b, \(\frac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}\) + \(\frac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
= \(\frac{\sqrt{3}\left(1+\sqrt{\sqrt{3}+1}\right)+\sqrt{3}\left(1-\sqrt{\sqrt{3}+1}\right)}{\left(1-\sqrt{\sqrt{3}+1}\right)\left(1+\sqrt{\sqrt{3}+1}\right)}\)
= \(\frac{\sqrt{3}+\sqrt{3\sqrt{3}+3}+\sqrt{3}-\sqrt{3\sqrt{3}+3}}{1^2-\left(\sqrt{\sqrt{3}+1}\right)^2}\)
= \(\frac{2\sqrt{3}}{1-\sqrt{3}-1}\)
= \(\frac{2\sqrt{3}}{-\sqrt{3}}\)
= -2
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TM
22 tháng 3 2022 lúc 8:36
Thực hiện các phép tính :
1. \(A=\sqrt{2-\sqrt{3}}\sqrt{2+\sqrt{2-\sqrt{3}}}\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{3}}}}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{3}}}}}\)
2. \(B=\left(\dfrac{1}{1+\sqrt{2}}+\dfrac{2}{2+\sqrt{3}}+...+\dfrac{1}{20+\sqrt{21}}\right)\cdot2022\)
Giải chi tiết giúp mình ạ
1:
\(A=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{2-\sqrt{3}}}\cdot\sqrt{2^2-\left(2+\sqrt{2-\sqrt{3}}\right)}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{2-\sqrt{3}}}\cdot\sqrt{2-\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{4-2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
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thực hiện phép tính:
1, dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}-dfrac{5-2sqrt{5}}{2sqrt{5}-4}
2,dfrac{sqrt{3}+1}{sqrt{3}-1}+dfrac{sqrt{3}-1}{sqrt{3}+1}
3,dfrac{2sqrt{8}-sqrt{12}}{sqrt{18}-sqrt{48}}+dfrac{sqrt{5}+sqrt{27}}{sqrt{30}-sqrt{2}}
4,dfrac{3-sqrt{3}}{2sqrt{3}-1}+dfrac{3+sqrt{3}}{2sqrt{3}-1}
5,dfrac{2sqrt{3}-4}{sqrt{3}-1}+dfrac{2sqrt{2}-1}{sqrt{2}-1}-dfrac{1+sqrt{6}}{sqrt{2}+sqrt{3}}
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thực hiện phép tính:
1, \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2,\(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
3,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}+\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)
4,\(\dfrac{3-\sqrt{3}}{2\sqrt{3}-1}+\dfrac{3+\sqrt{3}}{2\sqrt{3}-1}\)
5,\(\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
1: \(=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)
2: \(=\dfrac{4+2\sqrt{3}+4-2\sqrt{3}}{2}=\dfrac{8}{2}=4\)
4: \(=\dfrac{-3+5\sqrt{3}}{11}+\dfrac{3+5\sqrt{3}}{11}=\dfrac{10\sqrt{3}}{11}\)
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D sqrt{94-42sqrt{5}}-sqrt{94+42sqrt{5}}
A sqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
C dfrac{2sqrt{4-sqrt{5+sqrt{21+sqrt{80}}}}}{sqrt{10}-sqrt{2}}
F sqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{3}}}}
B dfrac{1}{sqrt{1}+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+...+dfrac{1}{sqrt{n-1}+sqrt{n}}
E dfrac{1}{sqrt{1}-sqrt{2}}-dfrac{1}{sqrt{2}-sqrt{3}}+dfrac{1}{sqrt{3}-sqrt{4}}-...-dfrac{1}{sqrt{24}-sqrt{25}}
Đọc tiếp
D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
F = \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
B = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}\)
E = \(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{24}-\sqrt{25}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{6+\sqrt{20}}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)}{10-2}\)
C = \(\dfrac{2\sqrt{30-10\sqrt{5}}+2\sqrt{6-2\sqrt{5}}}{8}\)
C = \(\dfrac{2\sqrt{\left(5-\sqrt{5}\right)^2}+2\sqrt{\left(\sqrt{5}-1\right)^2}}{8}\)
C = \(\dfrac{2\left(5-\sqrt{5}\right)+2\left(\sqrt{5}-1\right)}{8}\)
C = \(\dfrac{10-2\sqrt{5}+2\sqrt{5}-2}{8}\) = \(\dfrac{8}{8}\) = \(1\)
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D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
D = \(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)
D = \(7-3\sqrt{5}-\left(7+3\sqrt{5}\right)\) = \(7-3\sqrt{5}-7-3\sqrt{5}\)
D = \(-6\sqrt{5}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\) = \(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{5}+1}\) = \(\sqrt{1}=1\)
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a) frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}+frac{12}{sqrt{6}-3}-sqrt{6}b)frac{sqrt{3}+sqrt{2}-1}{2+sqrt{6}}+frac{sqrt{2}-sqrt{3}}{sqrt{2}+1}left(frac{sqrt{3}}{2-sqrt{6}}+frac{sqrt{3}}{2+sqrt{6}}right)-frac{1}{sqrt{2}}c) left(frac{2}{sqrt{3}-1}+frac{3}{sqrt{3}-2}+frac{15}{3-sqrt{3}}right)frac{1}{sqrt{3}+5}d) frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+...+frac{1}{sqrt{99}+sqrt{100}}
Đọc tiếp
a) \(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}+\frac{12}{\sqrt{6}-3}-\sqrt{6}\)b)\(\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{\sqrt{3}}{2-\sqrt{6}}+\frac{\sqrt{3}}{2+\sqrt{6}}\right)-\frac{1}{\sqrt{2}}\)c) \(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right)\frac{1}{\sqrt{3}+5}\)d) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
Làm mất căn mẫu và thu gọn
1) dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}+dfrac{5-2sqrt{5}}{2sqrt{5}-4}
2) left(1-dfrac{5+sqrt{5}}{1+sqrt{5}}right)left(dfrac{5-sqrt{5}}{1-sqrt{5}}-1right)
3) left(dfrac{3sqrt{125}}{15}-dfrac{10-4sqrt{5}}{sqrt{5}-2}right)dfrac{1}{sqrt{5}}
4) dfrac{1}{1+sqrt{2}}-dfrac{1}{1-sqrt{2}}
5) dfrac{1}{3+sqrt{5}}-dfrac{1}{sqrt{5}-3}
6) dfrac{sqrt{3}+sqrt{2}}{sqrt{3}-sqrt{2}}+dfrac{sqrt{2}-sqrt{3}}{sqrt{2}+sqrt{3}}
7) dfrac{4}{1-sqrt{3}}+dfrac{sqrt{3}-1}{sqrt{3}+1}
8) dfrac{s...
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Làm mất căn mẫu và thu gọn
1) \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2) \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
3) \(\left(\dfrac{3\sqrt{125}}{15}-\dfrac{10-4\sqrt{5}}{\sqrt{5}-2}\right)\dfrac{1}{\sqrt{5}}\)
4) \(\dfrac{1}{1+\sqrt{2}}-\dfrac{1}{1-\sqrt{2}}\)
5) \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{\sqrt{5}-3}\)
6) \(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\)
7) \(\dfrac{4}{1-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
8) \(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3}{\sqrt{2}-1}\)
9) \(\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}+1}-1}-\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}+1}+1}\)
10) \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
11) \(\dfrac{5}{1+\sqrt{6}}-\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}\)
12) \(\dfrac{5}{3-\sqrt{7}}-\dfrac{3}{\sqrt{2}+\sqrt{3}}+\dfrac{-1}{\sqrt{2}-1}\)
Giúp em giải với ạ! Help me~!
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
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