A= \(\dfrac{1}{5^2}\)+\(\dfrac{2}{5^3}\)+\(\dfrac{3}{5^4}\)+.....+\(\dfrac{n}{5^{n+1}}\)+......+\(\dfrac{11}{5^{12}}\) với n\(\in\)N.chứng minh A<\(\dfrac{1}{16}\)
A= \(\dfrac{1}{5^2}\)+\(\dfrac{2}{5^3}\)+\(\dfrac{3}{5^4}\)+.....+\(\dfrac{n}{5^{n+1}}\)+......+\(\dfrac{11}{5^{12}}\) với n\(\in\)N.chứng minh A<\(\dfrac{1}{16}\)
Ta có :
\(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+.............+\dfrac{n}{5^{n+1}}+.....+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{3^3}+........+\dfrac{n}{5^n}+..........+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+.....+\dfrac{n}{5^n}+....+\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5^2}+\dfrac{2}{5^3}+.....+\dfrac{n}{5^{n+1}}+........+\dfrac{11}{5^{12}}\right)\)\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+.........+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+.......+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\rightarrowđpcm\)
Cho A= \(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{11}{5^{12}}\) với n\(\in N\)
Chứng minh rằng a < \(\dfrac{1}{16}\)
Ta có: \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+...+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+...+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)
\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\)
⇒5A=15+252+...+11511⇒5A=15+252+...+11511
⇒5A−A=15+152+...+1511−11512⇒5A−A=15+152+...+1511−11512
⇒4A=15+152+...+1511−11512⇒4A=15+152+...+1511−11512
⇒20A=1+15+...+1510−11511⇒20A=1+15+...+1510−11511
⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)
⇒16A=1−12511+11512<1⇒16A=1−12511+11512<1
⇒A<116⇒A<116
Câu 1: Tìm a để \(\dfrac{5a-17}{4a-23}\) có giá trị lớn nhất.
Câu 2: Cho \(\dfrac{m}{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1998}\) ; m, n \(\in N\) . CMR m \(⋮\) 1999
Câu 3: CMR \(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}>\dfrac{5}{8}\)
Câu 4: CMR \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{n}{5^{n+1}}+...+\dfrac{11}{5^{12}}< \dfrac{1}{16}\) với n là STN.
Giúp mk với !
cau 1
de a dat gia tri lon nhat suy ra5a-17/4a-23 lon nhat
suy ra 4a-23 phai nho nhat khac 0 va la so nguyen duong
suy ra 4a-23=1
suy ra 4a=1+23=24
suy ra a=24 chia 4=6
vay de a nho nhat thi a=6
Câu 1: Tìm a để \(\dfrac{5a-17}{4a-23}\) có giá trị lớn nhất.
Câu 2: Cho \(\dfrac{m}{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1998}\) ; m, n \(\in N\) . CMR m \(⋮\) 1999
Câu 3: CMR \(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}>\dfrac{5}{8}\)
Câu 4: CMR \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{n}{5^{n+1}}+...+\dfrac{11}{5^{12}}< \dfrac{1}{16}\) với n là STN.
Giúp mk với !
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
tìm số tự nhiên thỏa mãn điều kiện
\(2\cdot2^2+3\cdot2^3+4\cdot2^4+........+n\cdot2^n=2^{n+11}\)
rút gọn : \(A=\left(\dfrac{2}{5}-\dfrac{5}{2}+\dfrac{1}{10}\right):\left(\dfrac{5}{2}-\dfrac{2}{3}+\dfrac{1}{12}\right)\)
tính:\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+......+\dfrac{1}{2017}}{\dfrac{2016}{1}+\dfrac{2003}{2}+\dfrac{2002}{3}+.......+\dfrac{1}{2016}}\)
CMR :\(5a+2b⋮13\Leftrightarrow9a+b⋮13\left(a,b\in Z\right)\)
tính GTBT:
a)a.\(\left(\dfrac{-3}{2}\right)+a.\dfrac{1}{4}-a.\dfrac{5}{6}vớia=\dfrac{3}{5}\)
b)\(\dfrac{2}{5}.b-\dfrac{1}{3}.b+b.\left(\dfrac{-1}{2}\right)vớib=\dfrac{6}{13}\)
c)c.\(\dfrac{5}{6}+\dfrac{3}{4}.c-\dfrac{11}{12}.c\) với c=\(\dfrac{2019}{2020}\)
a) Ta có: \(a\left(-\dfrac{3}{2}\right)+a\cdot\dfrac{1}{4}-a\cdot\dfrac{5}{6}\)
\(=a\left(-\dfrac{3}{2}+\dfrac{1}{4}-\dfrac{5}{6}\right)\)
\(=a\left(\dfrac{-18}{12}+\dfrac{3}{12}-\dfrac{10}{12}\right)\)
\(=a\cdot\dfrac{-25}{12}\)(1)
Thay \(a=\dfrac{3}{5}\) vào biểu thức (1), ta được:
\(\dfrac{3}{5}\cdot\dfrac{-25}{12}=\dfrac{-75}{60}=\dfrac{-5}{4}\)
a) A = \(\dfrac{19}{9}-\dfrac{-4}{11}-\dfrac{2}{-3}\)
b) B = \(\dfrac{-5}{6}+\dfrac{-7}{12}-\dfrac{1}{-5}\)
c) C = \(\dfrac{-3}{4}+\dfrac{11}{3}-\dfrac{-5}{18}\)
d) D = \(\dfrac{-19}{3}-\dfrac{4}{5}-\dfrac{-2}{3}\)
a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)
b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)
c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)
d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)
Chứng minh: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2015^2}+\dfrac{1}{2015}\)
Cho biểu thức: A=\(\dfrac{2}{n-1}\left(n\in Z\right).\)Tìm tất cả các giá trị nguyên của n để A là số nguyên.
Không quy đồng mẫu hãy so sánh a và B biết; A=\(\dfrac{12}{5^{2012}}+\dfrac{18}{5^{2013}}\); B=\(\dfrac{18}{5^{2012}}+\dfrac{12}{5^{2013}}\)
2) Để A là nguyên thì n - 1 là ước nguyên của 2
\(n-1=1\Rightarrow n=2\)
\(n-1=2\Rightarrow n=3\)
3) Ta gọi M là \(\dfrac{12}{5^{2012}}\)
\(M=\dfrac{5.12}{5^{2012}.5}=\dfrac{60}{5^{2013}}\)
\(\Rightarrow\) \(A=\dfrac{60}{5^{2013}}+\dfrac{18}{5^{2013}}=\dfrac{78}{5^{2013}}\)
Ta gọi Q là \(\dfrac{18}{5^{2012}}\)
\(Q=\dfrac{18}{5^{2012}}=\dfrac{18.5}{5^{2012}.5}=\dfrac{90}{5^{2013}}\)
\(\Rightarrow\) \(B=\dfrac{90}{5^{2013}}+\dfrac{12}{5^{2013}}=\dfrac{102}{5^{2013}}\)
\(\dfrac{90}{5^{2013}}< \dfrac{102}{5^{2013}}\Rightarrow A< B\)
Ai thấy đúng thì ủng hộ mink, thấy sai góp ý nha !!!