\(=1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=2022-\dfrac{1}{2022}=\dfrac{4088483}{2022}\)
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
Tính:1) dfrac{1}{3-2sqrt{2}}+dfrac{1}{2+sqrt{5}}2) dfrac{1}{3-2sqrt{2}}-dfrac{1}{3+2sqrt{2}}3) dfrac{1}{sqrt{5}-sqrt{7}}+dfrac{2}{1-sqrt{7}}4) dfrac{1}{5+2sqrt{6}}-dfrac{1}{5-2sqrt{6}}5) -dfrac{1}{sqrt{2}-sqrt{3}}-dfrac{3}{sqrt{18}+2sqrt{3}}
Đọc tiếp
Tính:
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{2+\sqrt{5}}\)
2) \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
3) \(\dfrac{1}{\sqrt{5}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
4) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\)
5) \(-\dfrac{1}{\sqrt{2}-\sqrt{3}}\)\(-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
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}}
Đọc tiếp
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) \(\dfrac{\sqrt{5}+\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\dfrac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}\)
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}
Đọc tiếp
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) \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
Tìm GTNN bt:A=\(\dfrac{2020x+2021\sqrt{1-x^2}+2022}{\sqrt{1-x^2}}\)
Tính:1) dfrac{1}{1+sqrt{5}}+dfrac{1}{sqrt{5}-1}2) dfrac{1}{sqrt{5}+sqrt{3}}-dfrac{1}{sqrt{5}-sqrt{3}}3) dfrac{1}{sqrt{2}-sqrt{3}}-dfrac{1}{sqrt{3}+sqrt{2}}4) dfrac{1}{3+sqrt{5}}+dfrac{1}{sqrt{5}-3}5) dfrac{1}{sqrt{2}-sqrt{6}}-dfrac{1}{sqrt{6}+sqrt{2}}LM CHI TIẾT GIÚP MK NHÉ
Đọc tiếp
Tính:
1) \(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}\)
2) \(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)
3) \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
4) \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{\sqrt{5}-3}\)
5) \(\dfrac{1}{\sqrt{2}-\sqrt{6}}-\dfrac{1}{\sqrt{6}+\sqrt{2}}\)
LM CHI TIẾT GIÚP MK NHÉ
\(\dfrac{\sqrt{1+\dfrac{2\sqrt{2}}{3}}+\sqrt{1-\dfrac{2\sqrt{2}}{3}}}{\sqrt{1+\dfrac{2\sqrt{2}}{3}}-\sqrt{1-\dfrac{2\sqrt{2}}{3}}}=2\)
19 tháng 9 2021 lúc 21:42
Cdfrac{1}{1+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+....dfrac{1}{sqrt{99}+sqrt{100}}Cdfrac{1left(1-sqrt{2}right)}{left(1+sqrt{2}right)left(1-sqrt{2}right)}+dfrac{1left(sqrt{2}-sqrt{3}right)}{left(sqrt{2}-sqrt{3}right)left(sqrt{2}+sqrt{3}right)}+dfrac{1left(sqrt{3}-sqrt{4}right)}{left(sqrt{3}-sqrt{4}right)left(sqrt{3}+sqrt{4}right)}+........dfrac{1left(sqrt{99}-sqrt{100}right)}{left(sqrt{99}-sqrt{100}right)left(sqrt{99}+sqrt{100}right)}Cdfrac{1-sqrt{2}}{1-2}+dfrac{sqrt{2}-sqr...
Đọc tiếp
\(C=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+....\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(C=\dfrac{1\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}+\dfrac{1\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{1\left(\sqrt{3}-\sqrt{4}\right)}{\left(\sqrt{3}-\sqrt{4}\right)\left(\sqrt{3}+\sqrt{4}\right)}+........\dfrac{1\left(\sqrt{99}-\sqrt{100}\right)}{\left(\sqrt{99}-\sqrt{100}\right)\left(\sqrt{99}+\sqrt{100}\right)}\)
\(C=\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+.....+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(C=\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{3}-\sqrt{4}}{-1}+......+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\)
\(C=-\left(1-\sqrt{2}\right)-\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{4}\right)-......-\left(\sqrt{99}-\sqrt{100}\right)\)
\(C=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-......-\sqrt{99}+\sqrt{100}\)
\(C=-1+\sqrt{100}\)
\(C=10-1=9\)
a)\(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\dfrac{1}{\sqrt{3}+5}\)
b)\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+3}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
CMR:
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)là 1 số hữu tỉ