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1. Bài giải:
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\)
\(\Rightarrow\frac{1}{2}A=A-\frac{1}{2}A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1000}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\right)\)
\(\Rightarrow\frac{1}{2}A=1-\frac{1}{1002}=\frac{1001}{1002}\Rightarrow A=\frac{2002}{1002}=\frac{1001}{501}\)
Vậy \(A=\frac{1001}{501}\)
Ta có : \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2019}=B\)
\(\Rightarrow A-B-1=-1\)
\(\Rightarrow\left(A-B-1\right)^{2019}=-1\)
\(a,=\left(\frac{9}{16}-\frac{10}{16}+\frac{12}{16}\right):\frac{11}{32}\)
\(=\frac{11}{16}:\frac{11}{32}\)
\(=\frac{11}{16}.\frac{32}{11}\)
\(=2\)
\(C=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.........\frac{2499}{2500}\)
\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}......\frac{49.51}{50^2}\)
\(=\frac{2.3.4....49}{3.4.5....50}.\frac{4.5.6....51}{3.4.5....50}\)
\(=\frac{1}{25}.17=\frac{17}{25}\)
\(a)\) \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{1000}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{999}{1000}\)
\(A=\frac{1.2.3.....999}{2.3.4.....1000}\)
\(A=\frac{1}{1000}.\frac{2.3.4.....999}{2.3.4.....999}\)
\(A=\frac{1}{1000}\)
Vậy \(A=\frac{1}{1000}\)
a) để 5/n-1 là số nguyên thì 5 chia hết cho n-1
=> n-1 thuộc Ư(5)=( 1, -1, 5, -5)
ta có
n-1=1=>n=2
n-1=-1=>n=0
n-1=5=>n=6
n-1=-5=>n=-4
mà n là số tự nhiên => n thuộc 2,0,6
máy mik bị lỗi bàn phím nên phải gõ ngoặc khác thay thế TvT, sorry nghen
b) M=(1-1000/2016) *...*(1-2016/2016)*(1-2017/2016)
=>M=(1-1000/2016)*.....*0*(1-2017/2016)
=>M=0
Bài 1:
\(\Leftrightarrow-\dfrac{5}{7}:x=-\dfrac{7}{18}-\dfrac{1}{6}=\dfrac{-7}{18}-\dfrac{3}{18}=\dfrac{-10}{18}=\dfrac{-5}{9}\)
=>x=5/9:5/7=7/9
Bài 2:
a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}=\dfrac{1000}{2}=500\)
b: \(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\)
\(=-\dfrac{1}{1000}\)
\(A-B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}-\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)\)
\(A-B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)+\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)-\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)\)
\(A-B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)+\left[\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)-\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)\right]\)
\(A-B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)-0\)
\(A-B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\)
\(\text{Thay }A-B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\text{ ta có : }\)
\(\left(A-B-1\right)^{1000}=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}-1\right)^{1000}\)
\(=\left(1-1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)^{1000}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)^{1000}\)