tim cac so nguyen a sao cho
(a2-1).(a2-4).(a2-7).(a2-10)>0
cho 10 so nguyen : a1;a2;a3;...a10 trong do moi so bang 1 hoac -1 Hoi tong S=a1.a2+a2.a3+a3.a4+...+a10.a1 co nhan gia tri bang 0 duoc khong? Vi sao? Cac ban giup minh nhe
Ban diep tim duoc hai so Nguyen STN (a2) = hai lam STH (a) nhung STH tru di 10 =STN tru di 5 (tuc a-10=a2- 5 ). tim hai so do
cho 7 so nguyen a1;a2;...;a7.Viet cac so nguyen do theo 1 thu tu khac duoc b1;b2;b3;...;b7. c/m rang tich so (a1-b1).(a2-b2)...(a7-b7) chia het cho 2
Tim STN n lon nhat sao cho so 2015 bang tong cua n so a1,a2,a3,...,an trong do tat ca cac so a1,a2,a3,...,an deu la hop so
tim so tu nhien n lon nhat sao cho so 2015 bang tong cua n so a1, a2,.....,an trong do tat ca cac so a1, a2, a3, .....an deu la hop so
tim so tu nhien n lon nhat sao cho so 2015 bang tong cua n so a1, a2,.....,an trong do tat ca cac so a1, a2, a3, .....an deu la hop so
Cho cac so nguyen duong a1,a2,...,a2017 thoa man 1/a1+1/a2+...+1/a2017=1009.chung minh co it nhat hai trong 2017 số trên bang nhau
Giả sử không có 2 số nào bằng nhau. Coi \(a_1>a_2>a_3>...>a_{2016}>a_{2017}\)
Do \(a_1;a_2;...;a_{2017}\in Z_+\)
\(\Rightarrow\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_{2017}}\le\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}< 1+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}=1009\)( Vô lý)
Do đó có ít nhất 2 số bằng nhau.
Bai 4:Cho day so co quy luat sau:
a1=1+2
a2=3+4+5
a3=6+7+8+9
a4=10+11+12+13+14
a,Tim a100
b,Tinh tong a1+a2+...+a100
cho 5 so nguyen phan biet a1,a2,a3,a4,a5.Xet tich:P=(a1-a2)*(a1-a3)*(a1-a4)*(a1-a5)*(a2-a3)*(a2-a4)*(a2-a5)*(a3-a4)*(a3-a5)*(a4-a5).Chung minh P chia het cho 288
Cho a,b,c>0 a2+b2+c2=3 Cmr: 1/(a+b) + 1/(b+c) + 1/(c+a) ≥ 4/(a2+7) + 4/(b2+7) + 4/(c2+7)
Ta có:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}\ge\dfrac{4}{a+2b+c}\ge\dfrac{4}{\dfrac{a^2+1}{2}+b^2+1+\dfrac{c^2+1}{2}}=\dfrac{8}{b^2+7}\)
Tương tự
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}\ge\dfrac{8}{a^2+7}\)
\(\dfrac{1}{b+c}+\dfrac{1}{a+c}\ge\dfrac{8}{c^2+7}\)
Cộng vế:
\(2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{8}{a^2+7}+\dfrac{8}{b^2+7}+\dfrac{8}{c^2+7}\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{4}{a^2+7}+\dfrac{4}{b^2+7}+\dfrac{4}{c^2+7}\)
Dấu "=" xảy ra khi \(a=b=c=1\)