Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
Giải:
Tên tam giác |
Tên 3 đỉnh |
Tên 3 góc |
Tên 3 cạnh |
ABI |
A,B,I |
|
AB, BI, IA |
AIC |
A,I,C |
|
AI, IC, CA |
ABC |
A,B,C |
|
AB, BC, CA |
15% của 3500 là : 525 .
6% của 140 là : 8,4 .
8% của 134 là : 10,72 .
40% của 1320 là : 528 .
15% của 3500 = 3500.15%=525
6% của 140 = 140.6%=8.4
8% của 134 = 134.8%= 10.72
40% của 1320 = 1320.40%= 528
Đặt A=\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)
=(\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\))+(\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\))+...+\(\dfrac{1}{70}\)
Nhận xét:
\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\)>\(\dfrac{1}{20}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{1}{20}\)=\(\dfrac{10}{20}\)=\(\dfrac{1}{2}\)
\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\)>\(\dfrac{1}{30}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{30}\)=\(\dfrac{10}{30}\)=\(\dfrac{1}{3}\)
\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\)>\(\dfrac{1}{60}\)+\(\dfrac{1}{60}\)+...+\(\dfrac{1}{60}\)=\(\dfrac{30}{60}\)=\(\dfrac{1}{2}\)
=>A>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)+\(\dfrac{1}{61}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)=\(\dfrac{4}{3}\)
=>A>\(\dfrac{4}{3}\)
Vậy: \(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{4}{3}\) (ĐPCM)
Thấy đúng cho 1 tick và 1 follow nha!
Chúc bạn học tốt!