Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
Rút gọn:
\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{5}-2}{5+2\sqrt{5}}-\dfrac{1}{2+\sqrt{5}}+\dfrac{1}{\sqrt{5}}\)
\(b.\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(c.\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+3}\)
Rút gọn:
C=\(\dfrac{3+\sqrt{5}}{2\sqrt{5}+\sqrt[]{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
D=\(\dfrac{1+\sqrt{\dfrac{3}{2}}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\sqrt{\dfrac{3}{2}}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}\)
Rút gọn
A=\(\sqrt{1^2+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1^2+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+......+\sqrt{1^2+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
HELP ME,PLS
C=\(\sqrt{\dfrac{\sqrt{2}-\sqrt{\dfrac{3}{2}}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}+}\sqrt{\dfrac{\sqrt{2}+\sqrt{\dfrac{3}{2}}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}}\)
Thực hiện phép tính:
a) \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)
b) \(B=\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right).\dfrac{1}{\sqrt{3}+5}\)
c) \(C=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
Tính:
Adfrac{sqrt{10}-sqrt{2}}{sqrt{5}-1}-dfrac{2-sqrt{2}}{sqrt{2}-1}
Bleft(1+dfrac{2+sqrt{2}}{1+sqrt{2}}right)cdotleft(1-dfrac{2-sqrt{2}}{1-sqrt{2}}right)
Cleft(dfrac{sqrt{6}-sqrt{2}}{1-sqrt{3}}right):dfrac{1}{sqrt{2}-sqrt{3}}
Ddfrac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}:dfrac{1}{sqrt{6}}
Edfrac{3+2sqrt{3}}{sqrt{3}}+dfrac{2+sqrt{2}}{1+sqrt{2}}-dfrac{1}{2-sqrt{3}}
Đọc tiếp
Tính:
\(A=\dfrac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}-\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(B=\left(1+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\right)\cdot\left(1-\dfrac{2-\sqrt{2}}{1-\sqrt{2}}\right)\)
\(C=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{2}-\sqrt{3}}\)
\(D=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{6}}\)
\(E=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
a) \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{19}+\sqrt{20}}\)
b) \(\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
rút gọn
dfrac{3+2sqrt{3}}{sqrt{3}}+dfrac{2+sqrt{2}}{1+sqrt{2}}-dfrac{1}{2-sqrt{3}}
left(dfrac{sqrt{6}-sqrt{2}}{1-sqrt{3}}-dfrac{5}{sqrt{5}}right).left(sqrt{5}-sqrt{2}right)
dfrac{2}{sqrt{5}+1}-sqrt{dfrac{2}{3-sqrt{5}}}
left(dfrac{1}{2-sqrt{5}}+dfrac{2}{sqrt{5}+sqrt{3}}right)/dfrac{1}{sqrt{21-12sqrt{3}}}
dfrac{2}{sqrt{3}-1}-sqrt{dfrac{2}{6-3sqrt{3}}}
dfrac{sqrt{2}}{2sqrt{2}+sqrt{3+sqrt{5}}}
Đọc tiếp
rút gọn
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right)/\dfrac{1}{\sqrt{21-12\sqrt{3}}}\)
\(\dfrac{2}{\sqrt{3}-1}-\sqrt{\dfrac{2}{6-3\sqrt{3}}}\)
\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)