\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
Những câu hỏi liên quan
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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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}}
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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) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
GIÚP EM ĐI ẠTÍNH:frac{3-sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}{3-sqrt{3+sqrt{6+sqrt{3}}}}+frac{2+sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}{3+sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}frac{1+frac{sqrt{3}}{2}}{1+sqrt{1+frac{sqrt{3}}{2}}}+frac{1-frac{sqrt{3}}{2}}{1-sqrt{1-frac{sqrt{3}}{2}}}frac{1}{sqrt{frac{5}{13}}+sqrt{frac{5}{7}}+1}+frac{1}{sqrt{frac{7}{5}}+sqrt{frac{7}{13}}+1}+frac{1}{sqrt{1frac{6}{7}}+1+sqrt{2frac{3}{5}}}RÚT GỌNsqrt{left(x-1right)^2}-x với x lớn hơn 1GIẢI PHƯƠNG TRÌNHsqrt{x^2-3x+2}+sqrt{x^2-4x+3}0
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GIÚP EM ĐI Ạ
TÍNH:
\(\frac{3-\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3-\sqrt{3+\sqrt{6+\sqrt{3}}}}+\frac{2+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}\)
\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
\(\frac{1}{\sqrt{\frac{5}{13}}+\sqrt{\frac{5}{7}}+1}+\frac{1}{\sqrt{\frac{7}{5}}+\sqrt{\frac{7}{13}}+1}+\frac{1}{\sqrt{1\frac{6}{7}}+1+\sqrt{2\frac{3}{5}}}\)
RÚT GỌN
\(\sqrt{\left(x-1\right)^2}-x\) với x lớn hơn 1
GIẢI PHƯƠNG TRÌNH
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
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Tính giá trị biểu thức:text{a) }frac{1}{sqrt{1}+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+frac{1}{sqrt{4}+sqrt{5}}+...+frac{1}{sqrt{2010}+sqrt{2011}}text{b) }frac{1}{2sqrt{1}+1sqrt{2}}+frac{1}{3sqrt{2}+2sqrt{3}}+frac{1}{4sqrt{3}+3sqrt{4}}+...+frac{1}{121sqrt{120}+120sqrt{121}}text{c) }sqrt{1+frac{1}{1^2}+frac{1}{2^2}}+sqrt{1+frac{1}{2^2}+frac{1}{3^2}}+sqrt{1+frac{1}{3^2}+frac{1}{4^2}}+...sqrt{+1+frac{1}{2010^2}+frac{1}{2011^2}}
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Tính giá trị biểu thức:
\(\text{a) }\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{2010}+\sqrt{2011}}\)
\(\text{b) }\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{121\sqrt{120}+120\sqrt{121}}\)
\(\text{c) }\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...\sqrt{+1+\frac{1}{2010^2}+\frac{1}{2011^2}}\)
Tính
a)\(\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+....+\frac{1}{\sqrt{1936}+\sqrt{1935}}\)
b)\(\frac{1}{\sqrt{1}-\sqrt{1}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}-\frac{1}{\sqrt{4}+\sqrt{5}}+....-\frac{1}{\sqrt{8}-\sqrt{2}}\)
a)\(\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{1936}+\sqrt{1935}}=\)
\(\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\)\(+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right)\left(\sqrt{4}+\sqrt{3}\right)}+...\)\(+\frac{\sqrt{1936}-\sqrt{1935}}{\left(\sqrt{1936}-\sqrt{1935}\right)\left(\sqrt{1936}+\sqrt{1935}\right)}\)= \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{1936}-\sqrt{1935}\)= \(-1-\sqrt{1935}\)
b)đề hơi sai bạn ạ mẫu thức số một bằng 0 còn đâu sửa lại đề đi nhé sau đó trục căn thức tương tự như mk làm nha
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cảm ơn bạn nha mik ghi dề sai đề đúng là như thế này nè\(\frac{1}{\sqrt{1}-\sqrt{2}}\) bạn giải giúp mik lun đi mik cảm ơn b nhìu lắm
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trục căn thức ở mẫu :
a,frac{3}{sqrt{5}};frac{2sqrt{3}}{sqrt{2}};frac{a}{sqrt{b}};frac{x+1}{sqrt{x^2-1}}
b,frac{1}{sqrt{3}+sqrt{2}};frac{2}{2-sqrt{3}};frac{sqrt{2}+1}{sqrt{2}-1};frac{3sqrt{2}}{sqrt{3}+1}
c,frac{1}{1+sqrt{2}+sqrt{3}}
d,frac{1}{sqrt{2sqrt{3}-sqrt{2}}.sqrt{2}.sqrt{sqrt{2}+sqrt{3}}}
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trục căn thức ở mẫu :
a,\(\frac{3}{\sqrt{5}};\frac{2\sqrt{3}}{\sqrt{2}};\frac{a}{\sqrt{b}};\frac{x+1}{\sqrt{x^2-1}}\)
b,\(\frac{1}{\sqrt{3}+\sqrt{2}};\frac{2}{2-\sqrt{3}};\frac{\sqrt{2}+1}{\sqrt{2}-1};\frac{3\sqrt{2}}{\sqrt{3}+1}\)
c,\(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)
d,\(\frac{1}{\sqrt{2\sqrt{3}-\sqrt{2}}.\sqrt{2}.\sqrt{\sqrt{2}+\sqrt{3}}}\)
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
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câu c chắc là như này
\(\frac{1}{1+\sqrt{2}+\sqrt{3}}=1+\frac{1}{\sqrt{2}+\sqrt{3}}\) = \(1+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
= \(1+\frac{\sqrt{2}-\sqrt{3}}{2-3}=1+\frac{\sqrt{2}-\sqrt{3}}{-1}\) = \(1-\sqrt{2}+\sqrt{3}\)
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Tính:
a, \(\frac{2}{5+2\sqrt{6}}+\frac{20}{\sqrt{6}-1}+\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}-\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b,\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
b/ \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\)
\(=\sqrt{n+1}-1\)
Câu a quy đồng từ từ từ phải qua trái là ra
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1-a,\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
b,\(B=\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
Ta có :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
Ta có:
\(\frac{1}{\sqrt{x}+\sqrt{x-1}}=\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}=\sqrt{x}-\sqrt{x-1}\)
Do đó:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)
\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)
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thực hiện phép tính
a )\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{1680}+\sqrt{1681}}\)
b) \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
b/ Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}.\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{99}-\frac{1}{\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
Cả 2 câu là n tự nhiên khác 0 hết nhé
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a/ Ta có: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
Áp đụng vào bài toán được
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{1680}+\sqrt{1681}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{1681}-\sqrt{1680}\)
\(=\sqrt{1681}-\sqrt{1}=41-1=40\)
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1) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
2) \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(A=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=8-\frac{3+\sqrt{5}}{2}=\frac{13-\sqrt{5}}{2}\)
\(B=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{8+2\sqrt{15}+8-2\sqrt{15}}{2}=8\)
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