xac dinh gia tri cua so a1,a2........a2009 la cac so nguyen thoa man
a1+a2+..........+a2009=0
a1+a2=a3+a3..........+a2009=0
Cho dãy tỉ số: a1/a2=a2/a3=a3/a4=...=a2008/a2009
Cmr: a1/a2009=(a1+a2+a3+...+a2008/a2+a3+a4+...+a2009)^2008
cho a1\a2=a2\a3=a3\a4=...=a2008\a2009. chứng minh a1\a2009=(a1+a2+....+a2008\a2+a3+....+a2009)2008 nhanh hộ mik nha
Đặt: \(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}=...=\dfrac{a_{2008}}{a_{2009}}=t\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}=...=\dfrac{a_{2008}}{a_{2009}}=\dfrac{a_1+a_2+a_3+...+a_{2008}}{a_2+a_3+...+a_{2009}}=t\)
Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{a_1+a_2+...+a_{2008}}{a_2+a_3+...+a_{2009}}\right)^{2008}=t^{2008}\\\dfrac{a_1}{a_2}.\dfrac{a_2}{a_3}.\dfrac{a_3}{a_4}...\dfrac{a_{2008}}{a_{2009}}=t^{2008}=\dfrac{a_1}{a_{2009}}\end{matrix}\right.\Leftrightarrow\left(đpcm\right)\)
cho dãy tỉ số bằng nhau.
a1/a2=a2/a3=...=a2008/a2009.
CMR:a1/a2009=(a1+a2+a3+a4+...+a2008)/(a2+a3+a4+a5+...+a2009)
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
a) Cho \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\)
CMR:(\(\dfrac{a+b+c}{b+c+d}\))\(^3\)=\(\dfrac{a}{d}\)
b)Cho \(\dfrac{a1}{a2}\)=\(\dfrac{a2}{a3}\)=\(\dfrac{a3}{a4}\)=...=\(\dfrac{a2008}{a2009}\)
CMR: \(\dfrac{a1}{a2009}\)=(\(\dfrac{a1+a2+a3+...+a2008}{a2+a3+a4+...+a2009}\))\(^{2008}\)
c) Cho \(\dfrac{a}{2003}\)=\(\dfrac{b}{2004}\)=\(\dfrac{c}{2005}\)
CMR: 4(a-b)(b-c)=(c-a)\(^2\)
a: Đặt a/b=b/c=c/d=k
=>a=bk; b=ck; c=dk
=>a=bk; b=dk^2; c=dk
=>a=dk^3; b=dk^2; c=dk
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=k^3\)
\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)
=>\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
c: Đặt a/2003=b/2004=c/2005=k
=>a=2003k; b=2004k; c=2005k
4(a-b)(b-c)=(c-a)^2
=>4(2004k-2003k)(2005k-2004k)=(2005k-2003k)^2
=>4*k*k=(2k)^2(luôn đúng)
=>ĐPCM
a) Cho \(\frac{a}{b}\)= \(\frac{b}{c}\)=\(\frac{c}{d}\)
CMR:(\(\frac{a+b+c}{b+c+d}^{ }\))\(^3\)=\(\frac{a}{d}\)
b) Cho \(\frac{a1}{a2}\)=\(\frac{a2}{a3}\)=\(\frac{a3}{a4}\)=...=\(\frac{a2008}{a2009}\)
CMR:\(\frac{a1}{a2009}\)=(\(\frac{a1+a2+a3+...+a2008}{a2+a3+a4+...+a2009}\))\(^{2008}\)
c) Cho \(\frac{a}{2003}\)=\(\frac{b}{2004}\)=\(\frac{c}{2005}\)
CMR: 4(a-b)(b-c)=(c-a)\(^{^2}\)
Cho a1,a2,a3 .....an la moi so co gia tri = 1 hoac = (-1)
Biet a1 + a2 + a3 + ... +an = 0
C/m n chia hết cho 4
cho 20 so nguyen a1,a2,a3,...,a20 khac 0 thoa man ,a1>0 va tong cua 3 so lien nhau bat ki la so duong va tong 20 so da cho la so am chung minh rang a1*a14+a12*a14<a1*a12
sai đề : phải là: a1.a14+a14.a12<a1.a12 nếu thế thì giải như sau
Ta có : a1 + (a2 + a3 + a4) + … + (a11 + a12 + a13) + a14 + (a15 + a16 + a17) + (a18 + a19 + a20) < 0 ; a1 > 0 ; a2 + a3 + a4 > 0 ; … ; a11 + a12 + a13 > 0 ; a15 + a16 + a17 > 0 ; a18 + a19 + a20 > 0 => a20 < 0.
Cũng như vậy : (a1 + a2 + a3) + … + (a10 + a11 + a12) + (a13 + a14) + (a15 + a16 + a17) + (a18 + a19 + a20) < 0 => a13 + a14 < 0.
Mặt khác, a12 + a13 + a14 > 0 => a12 > 0.
Từ các điều kiện a1 > 0 ; a12 > 0 ; a14 < 0 => a1.a14 + a14a12 < a1.a12 [dpcm]
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