Chung minh rang:\(A=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+........+\dfrac{1}{20}\)\(>\dfrac{1}{2}\)
Chung minh
\(A=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+.....+\dfrac{1}{20}\)\(< \dfrac{1}{2}\)
Ta có :
\(A=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{1}{12}+\dfrac{1}{12}+...+\dfrac{1}{12}\left(6PS\right)\)
Mà\(\dfrac{1}{12}+\dfrac{1}{12}+...+\dfrac{1}{12}=6.\dfrac{1}{12}=\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
\(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{1}{2}\\ \dfrac{1}{10}>\dfrac{1}{12}\\ \dfrac{1}{12}=\dfrac{1}{12}\\ ...\\ \dfrac{1}{20}< \dfrac{1}{12}\)
⇒Cộng 2 vế, ta có:
\(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{6}{12}=\dfrac{1}{2}\)
Vậy A<\(\dfrac{1}{2}\)
Cho minh hoi bai nay vs:
Chung to S=\(\dfrac{1}{4}\)+\(\dfrac{2}{4}\)+...+\(\dfrac{10}{4}\)<14
Chung to S=\(\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}< 2\)
Chung to S=\(\dfrac{1}{10}+\dfrac{1}{4}+...+\dfrac{1}{100}>1\)
— S = 1/4 + 2/4 +...+10/4 (1)
= 1 + 1/4 + 2/4 +...+ 9/4 (2)
=> Lấy (2) trừ đi (1) ta được:
1 — 10/4 = —6/4
Vì 14 = 14/1 = 84/6 mà —6/4 < 84/6
Do đó S < 14
VÒNG 2
Bài 1: Mèo con nhanh nhẹn
\(\dfrac{1}{2}\) + \(\dfrac{1}{12}\) | 2 + \(\dfrac{1}{6}\) | \(\dfrac{1}{20}\) | 1 - \(\dfrac{1}{9}\) | |
\(\dfrac{1}{15}\) + \(\dfrac{2}{15}\) | \(\dfrac{1}{2}\) + \(\dfrac{2}{3}\) | \(\dfrac{7}{12}\) | \(\dfrac{4}{12}\) | |
\(\dfrac{9}{14}\)+ \(\dfrac{1}{14}\) | 1 + \(\dfrac{1}{6}\) | \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) | \(\dfrac{1}{3}\) - \(\dfrac{2}{9}\) | |
\(\dfrac{3}{2}\) + \(\dfrac{2}{3}\) | \(\dfrac{1}{5}\) | 1 - \(\dfrac{8}{9}\) | ||
\(\dfrac{5}{7}\) | 1 - \(\dfrac{2}{3}\) | \(\dfrac{1}{3}\) + \(\dfrac{5}{9}\) |
Tính giá trị của biểu thức:
A=\(\dfrac{1}{9}\).\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\).\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\).\(\dfrac{1}{12}\)+\(\dfrac{1}{12}\).\(\dfrac{1}{13}\)+\(\dfrac{1}{13}\).\(\dfrac{1}{14}\)+\(\dfrac{1}{14}\).\(\dfrac{1}{15}\)
Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)
\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}\)
\(=\dfrac{2}{45}\)
-Chúc bạn học tốt-
A = \(\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
= \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\)
= \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+...+\dfrac{1}{14}-\dfrac{1}{15}\)
= \(\dfrac{1}{9}-\dfrac{1}{15}\)
= \(\dfrac{2}{45}\)
a,Chung to rang\(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}\)>\(\dfrac{1}{12}\)
Ai nhanh tick
Đặt \(A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+\dfrac{1}{44}+...+\dfrac{1}{80}\)
\(=\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}\right)\)
Nhận xét:
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}\) \(=\dfrac{1}{3}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}>\dfrac{1}{80}+\dfrac{1}{80}+...+\dfrac{1}{80}\) \(=\dfrac{1}{4}\)
\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}>\dfrac{1}{12}\)
Vậy \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{1}{12}\) (Đpcm)
Chung minh rang : \(\dfrac{2}{x^2-1}+\dfrac{4}{x^2-4}+\dfrac{6}{x^2-9}+............+\dfrac{20}{x^2-100}\)chia het cho 11
Tìm x, biết:
a) \(\dfrac{-1}{10}\) + \(\dfrac{2}{5}\)x + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
b) \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\) : x= \(-\dfrac{1}{5}\)
c) \(-\dfrac{2}{3}\) : x + \(\dfrac{5}{8}\) = \(-\dfrac{7}{12}\)
a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\)\(x\) = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\) \(x\) = - \(\dfrac{3}{20}\)
\(x\) = - \(\dfrac{3}{20}\): \(\dfrac{2}{5}\)
\(x\) = - \(\dfrac{3}{8}\)
b, \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{8}{15}\)
\(x\) = \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))
\(x\) = - \(\dfrac{15}{16}\)
c, - \(\dfrac{2}{3}\): \(x\) + \(\dfrac{5}{8}\) = - \(\dfrac{7}{12}\)
\(\dfrac{2}{3}\): \(x\) = \(\dfrac{7}{12}\) + \(\dfrac{5}{8}\)
\(\dfrac{2}{3}\) : \(x\) = \(\dfrac{29}{24}\)
\(x\) = \(\dfrac{2}{3}\) : \(\dfrac{29}{24}\)
\(x\) = \(\dfrac{16}{29}\)
Xin mời các đại tỉ cao nhân giúp em :((( em xin trân trọng cảm ơn :)))
Tính hợp lí:
\(a,\dfrac{6}{21}-\dfrac{-12}{44}+\dfrac{10}{14}-\dfrac{1}{-4}-\dfrac{18}{33}\\ b,\dfrac{3}{7}.\left(-\dfrac{2}{5}\right).2\dfrac{1}{3}.20.\dfrac{19}{72}\)
6/21-(−12/44)+10/14−(1/(−4))−18/33
=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11
=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11
=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4
=−3/11+7/7+1/4=−3/11+7/7+1/4
=43/44
cho A =\(\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\)SO SÁNH A VỚI \(\dfrac{1}{2}\)
Ta có:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}>4\cdot\dfrac{1}{16}=\dfrac{1}{4}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}>4\cdot\dfrac{1}{20}=\dfrac{1}{5}\)
=>\(\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}>\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{9}{20}\)
=>A>\(\dfrac{1}{12}+\dfrac{9}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
=>\(A>\dfrac{1}{20}+\dfrac{9}{20}=\dfrac{1}{2}\)
Vậy...