Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
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
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
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=\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{1+\dfrac{1}{4^2}+\dfrac{1}{5^2}}+.....+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
\(\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
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:
\(S=\dfrac{1}{2\sqrt{1}+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
Tính giá trị của biểu thức \(P=\dfrac{4\left(x+1\right)x^{2018}-2x^{2017}+2x+1}{2x^2+3x}\) tại \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}\)
\(\dfrac{1}{\sqrt{1^3+2^3}}+\dfrac{1}{\sqrt{1^2+2^3+3^3}}+.....+\dfrac{1}{\sqrt{1^3+2^3}+....+2018^3}\)