\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{39}{19^2.20^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{39}{361.400}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{361}-\dfrac{1}{400}\)
\(=1-\dfrac{1}{400}< 1\)
Vậy A < 1
Tính bằng cách hợp lí :
a , \(\dfrac{1}{15}+\dfrac{9}{10}+\dfrac{14}{15}-\dfrac{11}{9}-\dfrac{20}{10}+\dfrac{1}{157}\)
b , \(\dfrac{1}{5}-\dfrac{-1}{3}+\dfrac{-1}{5}-\dfrac{2}{6}\)
c , \(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{2015\times2017}\)
d , \(\dfrac{5}{1\times3}+\dfrac{5}{3\times5}+...+\dfrac{5}{2015\times2017}\)
e , \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+...+\dfrac{1}{2016\times2017}\)
Cho A= \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{4026}\) , B = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+....+\dfrac{1}{4025}\). So sánh \(\dfrac{A}{B}\) với \(1\dfrac{2013}{2014}\)
Tính B=\(\left(1-\dfrac{2}{2.3}\right)\).\(\left(1-\dfrac{2}{3.4}\right).\left(1-\dfrac{2}{4.5}\right)......\left(1-\dfrac{2}{99.100}\right)\)
Tính:
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}\)
Lẹ nha mấy chế!!!~.~
\(A=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
Cho biểu thức:
A= (\(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\)) : (\(\dfrac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\))
a) Rút gọn A
b) Tính A khi x = \(\dfrac{3-2\sqrt{2}}{4}\)
cho a,b,c>0.cmr
\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
Cho R={3k-1| k∈R, -5≤ k ≤5}, S={x ∈R| \(3< \left|x\right|\le\dfrac{19}{2}\)}, T= {x∈R| 2x2-4x+2=0}. Tính \(R\cap S,S\cup T\),R\S
Cho a , b , c > 0 ; a + b + c = 3 . Chứng minh rằng \(\dfrac{a}{1+b^2}\) + \(\dfrac{b}{1+c^2}\) + \(\dfrac{c}{1+a^a}\) \(\ge\) \(\dfrac{3}{2}\)
Giúp mk vs
