\(=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}\)
=2017-1/2019
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Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
Rút gọn:
\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
a) \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{19}+\sqrt{20}}\)
b) \(\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
cau a cho x,y,z\(\ne\)0 thoa man x+y+z=0. CM: \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}|\) cau b tinh G=\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{1+\dfrac{1}{4^2}+\dfrac{1}{5^2}}+.....+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(\dfrac{1}{\sqrt{1^3+2^3}}+\dfrac{1}{\sqrt{1^2+2^3+3^3}}+.....+\dfrac{1}{\sqrt{1^3+2^3}+....+2018^3}\)
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
so sánh \(\sqrt{2019^2-1}-\sqrt{2018^2-1}\)và \(\dfrac{2.2018}{\sqrt{2019^2-1}+\sqrt{2018^2-1}}\)
Tính tổng:
\(S=\dfrac{1}{2\sqrt{1}+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
giải pt : \(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}+\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}+\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}+...+\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2020}}=11\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{5}-2}{5+2\sqrt{5}}-\dfrac{1}{2+\sqrt{5}}+\dfrac{1}{\sqrt{5}}\)
\(b.\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(c.\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+3}\)