bn chỉ cần trục căn thức từng cái thì sẽ có mẫu chung thôi
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Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
Rút gọn:
\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Tính tổng:
\(S=\dfrac{1}{2\sqrt{1}+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
pdfrac{2}{1-sqrt{2}}-dfrac{2}{1+sqrt{2}}
Qleft(dfrac{sqrt{6}-sqrt{2}}{1-sqrt{3}}-dfrac{5}{sqrt{5}}right)cdotleft(sqrt{5}-sqrt{2}right)
Rdfrac{2}{7+4sqrt{3}}+dfrac{2}{7-4sqrt{3}}
Sdfrac{2}{sqrt{5}+1}-sqrt{dfrac{2}{3-sqrt{5}}}
Tdfrac{4}{1-sqrt{3}}-dfrac{sqrt{15}+sqrt{3}}{1+sqrt{5}}
Uleft(dfrac{1}{2-sqrt{5}}+dfrac{2}{sqrt{5}+sqrt{3}}right):dfrac{1}{sqrt{21-12sqrt{3}}}
Vdfrac{2}{sqrt{3}-1}-sqrt{dfrac{2}{6-3sqrt{3}}}
Wdfrac{5sqrt{3}}{sqrt{3-sqrt{5}}-sqrt{3}}-dfrac{5sqrt{3}}{sqrt{3-sqrt{5}}+sqr...
Đọc tiếp
\(p=\dfrac{2}{1-\sqrt{2}}-\dfrac{2}{1+\sqrt{2}}\)
\(Q=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)\)
\(R=\dfrac{2}{7+4\sqrt{3}}+\dfrac{2}{7-4\sqrt{3}}\)
\(S=\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(T=\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
\(U=\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21-12\sqrt{3}}}\)
\(V=\dfrac{2}{\sqrt{3}-1}-\sqrt{\dfrac{2}{6-3\sqrt{3}}}\)
\(W=\dfrac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}-\sqrt{3}}-\dfrac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}+\sqrt{3}}\)
\(Y=\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
a) \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{19}+\sqrt{20}}\)
b) \(\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
\(S=\dfrac{\sqrt{2}-\sqrt{1}}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{25}-\sqrt{29}}{29+25}< \dfrac{2}{5}\)
Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
Tính tổng: \(S=\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{1+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{5}-2}{5+2\sqrt{5}}-\dfrac{1}{2+\sqrt{5}}+\dfrac{1}{\sqrt{5}}\)
\(b.\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(c.\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+3}\)