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b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
=>x=3/11+20/55=3/11+4/11=7/11
c: \(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-2}{98}-1\right)+\left(\dfrac{x-5}{95}-1\right)=\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)
\(\Leftrightarrow x-100=1\)
hay x=101
Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:
a) \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{1}{1.2}\\\dfrac{1}{6}=\dfrac{1}{2.3}\\\dfrac{1}{12}=\dfrac{1}{3.4}\\...\end{matrix}\right.\)
Vậy số thứ 100 của dãy là: \(\dfrac{1}{100.101}=\dfrac{1}{10100}\)
Tổng: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{6}=\dfrac{1}{\left(5.0+1\right)\left(5.1+1\right)}\\\dfrac{1}{66}=\dfrac{1}{\left(5.1+1\right)\left(5.2+1\right)}\\\dfrac{1}{176}=\dfrac{1}{\left(5.2+1\right)\left(5.3+1\right)}\\...\end{matrix}\right.\)
Vậy số thứ 100 của dãy là: \(\dfrac{1}{\left(5.99+1\right)\left(5.100+1\right)}=\dfrac{1}{248496}\)
Tổng: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{496.501}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{496.501}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}.\dfrac{500}{501}\)
\(=\dfrac{100}{501}\)
Bài 2: Tính:
a) \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(A=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(A=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(\Rightarrow A=\dfrac{100}{2}=50\)
Câu 3:
a: \(A=-\left|x-10\right|+2018< =2018\)
Dấu '=' xảy ra khi x=10
\(B=-\left(x+2\right)^2+1999< =1999\)
Dấu '=' xảy ra khi x=-2
b: \(A=\left(2x-8\right)^2+3>=3\)
Dấu '=' xảy ra khi x=4
\(B=\left|x^2-25\right|-2017>=-2017\)
Dấu '=' xảy ra khi x=5 hoặc x=-5
\(a.\)
\(\dfrac{3}{16}:\dfrac{?}{8}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{16}\cdot\dfrac{8}{?}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{2?}=\dfrac{3}{4}\)
\(\Leftrightarrow?=2\)
\(b.\)
\(\dfrac{1}{25}:-\dfrac{3}{?}=-\dfrac{1}{15}\)
\(\Leftrightarrow\dfrac{1}{25}\cdot\dfrac{-?}{3}=-\dfrac{1}{15}\)
\(\Leftrightarrow\dfrac{-?}{75}=-\dfrac{1}{15}\)
\(\Leftrightarrow?=\dfrac{75}{15}=5\)
\(c.\)
\(\dfrac{?}{12}:-\dfrac{4}{9}=-\dfrac{3}{16}\)
\(\Leftrightarrow\dfrac{?}{12}\cdot\dfrac{-9}{4}=-\dfrac{3}{16}\)
\(\Leftrightarrow\dfrac{-3?}{16}=-\dfrac{3}{16}\)
\(\Leftrightarrow?=1\)
Mk gọi ? = x nha
a) \(\dfrac{3}{16}:\dfrac{x}{8}=\dfrac{3}{4}\)
\(\dfrac{x}{8}=\dfrac{3}{16}:\dfrac{3}{4}\)
\(\dfrac{x}{8}=\dfrac{1}{4}\)
⇒\(x=\dfrac{1.8}{4}=2\)
b) \(\dfrac{1}{25}:\dfrac{-3}{x}=\dfrac{-1}{15}\)
\(\dfrac{-3}{x}=\dfrac{1}{25}:\dfrac{-1}{15}\)
\(\dfrac{-3}{x}=\dfrac{-3}{5}\)
⇒x=5
c) \(\dfrac{x}{12}:\dfrac{-4}{9}=\dfrac{-3}{16}\)
\(\dfrac{x}{12}=\dfrac{-3}{16}.\dfrac{-4}{9}\)
\(\dfrac{x}{12}=\dfrac{1}{12}\)
⇒x=1
\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)
\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)
=2022(1/2+1/3+...+1/2021+1/2022)
=>B/A=2022
a) Trước hết ta chứng minh \(a^2-1=\left(a-1\right)\left(a+1\right)\text{tự chứng minh }\)
Áp dụng bổ đề trên ta có:
\(-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right) =\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{100^2-1}{100^2}=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}=\dfrac{1\cdot2\cdot3^2\cdot...\cdot99^2\cdot100\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\dfrac{1\cdot101}{2\cdot100}>\dfrac{1}{2}\\ \Rightarrow A< -\dfrac{1}{2}\)
Gọi x là thương A:B cần tìm.Theo đề, ta có:
\(\left(\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{100.125}\right)x=\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{25.125}\)
Nhân 2 vế cho 100, ta có:
\(4\left(\dfrac{25}{1.26}+\dfrac{25}{2.27}+...+\dfrac{25}{100.125}\right)x=\dfrac{100}{1.101}+\dfrac{100}{2.102}+...+\dfrac{100}{25.125}\)
\(\Rightarrow4\left(1-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{100}-\dfrac{1}{125}\right)x=1-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+...+\dfrac{1}{25}-\dfrac{1}{125}\)
\(\Rightarrow4\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{125}\right)\right]x=\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)\)\(\Rightarrow4x=1\Rightarrow x=\dfrac{1}{4}\)
Vậy hiệu A:B là:\(\dfrac{1}{4}\)