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 < \(\dfrac{1}{16}\)
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<\(\dfrac{1}{16}\)
\(5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{11}{5^{11}}.\)
\(4A=5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}=B-\dfrac{11}{5^{12}}.\)
\(5B=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{10}}.\)
\(4B=5B-B=1-\dfrac{1}{5^{11}}\)
\(\Rightarrow4A=\dfrac{1}{4}\left(1-\dfrac{1}{5^{11}}\right)-\dfrac{1}{5^{12}}< \dfrac{1}{4}\Rightarrow A< \dfrac{1}{16}\)
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 :
\(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+.............+\dfrac{n}{5^{n+1}}+.....+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{3^3}+........+\dfrac{n}{5^n}+..........+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+.....+\dfrac{n}{5^n}+....+\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5^2}+\dfrac{2}{5^3}+.....+\dfrac{n}{5^{n+1}}+........+\dfrac{11}{5^{12}}\right)\)\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+.........+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+.......+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\rightarrowđpcm\)
a) Tìm số tự nhiên n sao cho P= 2.24n+1 + 1 là số nguyên tố
b) Cho P= \(\dfrac{1}{5^2}\) + \(\dfrac{2}{5^3}\) + \(\dfrac{3}{5^4}\) +....+ \(\dfrac{11}{5^{12}}\) . Chứng minh rằng: P < \(\dfrac{1}{16}\)
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 !
Cho A=\(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+....+\dfrac{10}{5^{10}}+\dfrac{11}{5^{11}}\). Chứng minh A<\(\dfrac{5}{16}\)
Cho \(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{11}{5^{11}}.\)
Chứng minh :\(A< \dfrac{5}{16}.\)
1, A= \(\dfrac{-3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
2, B= \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\)
3, C= \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
4, D= \(6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)
\(D=\dfrac{77}{39}+\dfrac{3}{2}\)
\(D=\dfrac{271}{78}\)
\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)
\(C=\dfrac{5}{2}-\dfrac{3}{2}\)
\(C=1\)
Bài 1 : Thực hiện phép tính
a , \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
b , \(\dfrac{-5}{11}.\dfrac{4}{13}+\dfrac{-5}{11}.\dfrac{9}{13}+3\dfrac{5}{11}\)
c , \(3^2-12.\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
Bài 2 : Cho đường thẳng xy , lấy điểm O thuộc đường thẳng xy . Trên tia Ox , lấy 2 điểm A , B sao cho OA = 3 cm , AB = 2 cm
a, Trong 3 điểm O , A , B điểm nào nằm giữa 2 điểm còn lại.
b, Có tất cả bao nhiêu tia ? nêu tên ?
c, Tính độ dài OB ?
Bài 3 : Tính A = \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{2021.2022}\)
bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8