chứng minh rằng \(\frac{91}{1\cdot4}+\frac{91}{4\cdot7}+\frac{91}{7\cdot10}+......+\frac{91}{88\cdot91}\)=30
chứng minh :91/1*4+91/4*7+61/7*10+.......91/88*91=30
nếu đúng mình tick luôn
Ta có \(\frac{91}{1.4}+\frac{91}{4.7}+\frac{91}{7.10}+...+\frac{91}{88.91}=\frac{91}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{88.91}\right)\)
\(=\frac{91}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{88}-\frac{1}{91}\right)\)
\(=\frac{91}{3}.\left(1-\frac{1}{91}\right)=\frac{91}{3}.\frac{90}{91}=30\)
9/1.4+9/4.7+9/7.10+…+9/88.91=30
=91/3. (3/1.4+3/4.7+3/7.10+…+3/88.91)
=91/3. (1-1/4+1/4-1/7+1/7-1/10+…+1/88-1/91)
=91/3*90/91=30
Chứng minh rằng: \(\frac{40^2+51^2+91^2}{79^2}=\frac{40^4+51^4+91^4}{79^4}\)
Cho \(S=\frac{1}{101}+\frac{1}{102}+...........+\frac{1}{130}\). Chứng minh rằng: \(\frac{1}{4}
C1:Chứng minh : \(\frac{91}{1.4}+\frac{91}{4.7}+\frac{91}{7.10}+.....+\frac{91}{88.91}=30\)
C2:Cho n là số tự nhiên có 2 chữ số .Tìm n biết n+4 và 2n đều là các số chính phương
a) \(\frac{91}{1.4}+\frac{91}{4.7}+...+\frac{91}{88.91}=\frac{91}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{88.91}\right)\)
\(=\frac{91}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{88}-\frac{1}{91}\right)=\frac{91}{3}\left(1-\frac{1}{91}\right)=\frac{91}{3}.\frac{90}{91}=30\left(\text{đpcm}\right)\)
Tính \(A=\frac{1}{3\cdot7}+\frac{1}{7\cdot11}+\frac{1}{11\cdot15}+...+\frac{1}{91\cdot95}\)
A=1/3.7+1/7.11+1/11.15+...+1/91.95
=>4A=4/3.7+4/7.11+4/11.15+...+4/91.95
4A=1/3-1/7+1/7-1/11+1/11-1/15+...1/91-1/95
4A=1/3-1/95
4A=92/285
A=92/285:4
A=23/285
Bạn chỉ cần lấy 1/3 trừ 1/95 là đc r, ko cần : 4 đâu
^^!
\(A=\frac{1}{1\cdot3}+\frac{2}{3\cdot7}+\frac{3}{7\cdot13}+...+\frac{10}{91\cdot111}\)
Ta có:
A=1/1.3+2/3.7+3/7.13+...+10/91.111
=>2A=2/1.3+4/3.7+6/7.13+...+20/91.111
=>2A=1-1/3+1/3-1/7+1/7-1/13+...+1/91-1/111
=>2A=1-1/111=110/111
=>A=55/111
Vậy A=55/111
OK!
CHỨNG TỎ RẰNG :
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{67\cdot70}< 1\)
\(A=\frac{1}{1.4}+\frac{1}{2.7}+...+\frac{1}{67.70}\)
\(3A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{67.70}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{70}\)
\(3A=1-\frac{1}{70}=\frac{69}{70}\)
\(A=\frac{69}{70}:3=\frac{23}{70}\)
vì \(\frac{23}{70}< 1\)
nên \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{67.70}< 1\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{67.70}\)
\(=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{67.70}\right)\)
\(=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{67.70}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{70}\)
\(=1-\frac{1}{70}\)
\(=\frac{69}{70}< 1\)
Chứng minh \(\frac{40^4+51^4+91^4}{79^4}=\frac{40^2+51^2+91^2}{79^2}\)
Chứng minh \(\frac{40^2+51^2+91^2}{79^2}=\frac{40^4+51^4+91^4}{79^4}\)
\(\frac{40^2+51^2+91^2}{79^2}\)= \(\frac{2\times\left(40^2+51^2+91^2\right)}{2\times79^2}\)
=> dpcm