A=\(\frac{\frac{1}{2018}+\frac{2}{2017}+\frac{3}{2016}+....+\frac{2017}{2}+\frac{2018}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2019}}\). Chứng minh rằng A là số nguyên
Mong mọi người giúp
\(A=\frac{1}{2017}+\frac{2}{2017^2}+\frac{3}{2017^3}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\). Chứng minh rằng : A < \(\frac{2017}{2016^2}\)
Tìm x biết
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)
\(\Leftrightarrow\)\(x+1=2019\)
\(\Leftrightarrow\)\(x=2019-1\)
\(\Leftrightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
Bài 1:Tìm số tự nhiên có 4 chữ số sao cho số đó vừa là số chính phương vừa là 1 lập phương
Bài 2: Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}+\frac{1}{2019}\)
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
Hãy so sánh A/B với 1/2018
\(\frac{A}{B}>\frac{1}{2018}\)
Cho \(A=1-\frac{2017}{2019}+\left(\frac{2017}{2019}\right)^2-\left(\frac{2017}{2019}\right)^3+...+\left(\frac{2017}{2019}\right)^{2018}\)
Chứng minh A không là số nguyên.
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\) \(B=\frac{1}{2018}+\frac{2}{2017}+\frac{3}{2016}+...+\frac{2017}{2}+\frac{2018}{1}\)
Tính \(\frac{A}{B}\)
ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
Tính P = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2019}}{\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+....\frac{1}{2018}}\)
Chứng minh rằng
a) Với mọi số nguyên dương n có \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+..+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
b) \(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}< \sqrt{2017}+\sqrt{2018}\)
Hộ mình vs
Câu b đề sai nha, bây giờ đặt \(a=\sqrt{2017},b=\sqrt{2018}\)
Ta có \(\frac{a^2}{b}+\frac{b^2}{a}< a+b\Leftrightarrow ab\left(\frac{a^2}{b}+\frac{b^2}{a}\right)< ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3< ab\left(a+b\right)\)(1)
Mà \(ab\left(a+b\right)\le\left(a^2-ab+b^2\right)\left(a+b\right)=a^3+b^3\)(2)
Từ (1), (2) => Sai
a) Ta có:
\(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{k+1-k}{\left(k+1\right)\sqrt{k}}=\frac{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}\)\(< \frac{2\sqrt{k+1}\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k+1}\sqrt{k}}=\frac{2}{\sqrt{k}}-\frac{2}{\sqrt{k+1}}\)
Cho k=1,2,....,n rồi cộng từng vế ta có:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< \left(\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}\right)+\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}\right)\)\(+\left(\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}\right)+....+\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n-1}}< 2\)
Tính :
a) \(\text{A}=\left(1\times2\right)^{-1}+\left(2\times3\right)^{-1}+...+\left(2014\times2015\right)^{-1}\).
b) \(\text{B}=\frac{2018+\frac{2017}{2}+\frac{2016}{3}+\frac{2015}{4}+...+\frac{2}{2017}+\frac{1}{2018}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2018}+\frac{1}{2019}}\).
Câu 1. Tính hợp lý giá trị các biểu thức sau :
a. A = ( 689 - 31 ) - ( 269 - 131 )
b. B = \(\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}+1\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}\right)-\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}+1\right)\)c. C = \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!