cho \(m,n\in Z\) sao cho
\(\dfrac{m}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{426}+\dfrac{1}{427}\)
cmr:\(m⋮641\)
cho m,n thuộc Z sao cho
\(\dfrac{m}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{426}+\dfrac{1}{427}\)
cmr m chia hết cho 641
1. Cho N=\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\)
CMR \(\dfrac{3}{5}< N< \dfrac{4}{5}\)
2. Cho M=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{29}{3^{29}}-\dfrac{30}{3^{30}}\)
CMR \(M< \dfrac{3}{16}\)
3. Cho Q=\(\dfrac{2}{3}+\dfrac{8}{9}+\dfrac{26}{27}+...+\dfrac{3^{2021}-1}{3^{2021}}\)
CMR \(Q>\dfrac{4041}{2}\)
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 !
Bài 1: cho a, b > 0 và a + b <= 1. CMR: \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}>=3\)
Bài 2: cho x, y, z >=0 thỏa mãn x + y + z >0. CMR: \(\dfrac{x}{4x+4y+z}+\dfrac{y}{4y+4z+x}+\dfrac{z}{4z+4x+y}< =\dfrac{1}{3}\)
Bài 3: cho x, y, z > 0 thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\)
Tìm GTNN của \(\dfrac{1}{\sqrt{2x^2+y^2+3}}+\dfrac{1}{\sqrt{2y^2+z^2+3}}+\dfrac{1}{\sqrt{2z^2}+x^2+3}\)
Bài 1 :
Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)
Theo BĐT Cô - Si dưới dạng engel ta có :
\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)
Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)
Cho \(n\in N\) sao, \(n\ge0\). C/m \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)
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 :\(5a+2b⋮13\Leftrightarrow9a+b⋮13\left(a,b\in Z\right)\)
cho x,y,z khác 0 thỏa mãn: \(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\). CMR: ( x+y+z ). \(\left(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\right)\)
sửa lại đề: CMR: ( x+y+z ).\(\left(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\right)=36\)
cho 3 số dương x,y,z thỏa mãn \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=6\)
CMR: \(\dfrac{1}{3x+3y+2z}+\dfrac{1}{3x+2y+3z}+\dfrac{1}{2x+3y+3z}\le\dfrac{3}{2}\)
Ta có :
\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(2x+y+z\right)+\left(2y+x+z\right)}\)(1)
Áp dụng BĐT \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\left(1\right)\le\dfrac{1}{4}\left(\dfrac{1}{x+y+x+z}+\dfrac{1}{y+x+y+z}\right)\le\dfrac{1}{4}\left(\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\right)\)
\(=\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)
tương tự với hai ông còn lại sau đó cộng lại ta được:
\(\Sigma\dfrac{1}{3x+3y+2z}\le\dfrac{24}{16}=\dfrac{3}{2}\)