so sánh A= \(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
và B=\(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
So sánh
A=\(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
và B=\(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
so sánh:
\(A=\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
\(B=\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
So sánh: \(A=\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
và \(B=\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
\(\frac{2}{60}+\frac{2}{63}+\frac{2}{63}+\frac{2}{66}+....+\frac{2}{117}+\frac{2}{120}+\frac{2}{2003}\)
So sánh
A=\(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
và B=\(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
So sánh
A=\(\frac{2}{60.63}+\frac{2}{63.66}+..+\frac{2}{117.120}+\frac{2}{2003}\)
B=\(\frac{5}{40.44}+\frac{5}{44.48}+..+\frac{5}{76.80}+\frac{5}{2003}\)
\(A=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}+\frac{2}{2003}\)
\(\text{Đặt }C=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}\)
\(C=\frac{2}{3}\left(\frac{3}{60\cdot63}+\frac{3}{63\cdot66}+...+\frac{3}{117\cdot120}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)\)
\(C=\frac{2}{3}\cdot\frac{1}{120}\)
\(C=\frac{1}{180}\)
\(\text{Thay }C=\frac{1}{180}\text{Ta có : }\) \(A=\frac{1}{180}+\frac{2}{2003}\)
\(B=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}+\frac{5}{2003}\)
\(\text{Đặt }D=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}\)
\(D=\frac{5}{4}\left(\frac{4}{40\cdot44}+\frac{4}{44\cdot48}+...+\frac{4}{76\cdot80}\right)\)
\(D=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)\)
\(D=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)\)
\(D=\frac{5}{4}\cdot\frac{1}{80}\)
\(D=\frac{1}{64}\)
\(\text{Thay }D=\frac{1}{64}\text{ Ta có : }B=\frac{1}{64}+\frac{5}{2003}\)
\(\text{Vì }A=\frac{1}{180}+\frac{2}{2003}\text{ , }B=\frac{1}{64}+\frac{5}{2003}\)
\(\text{Có : }\frac{1}{180}< \frac{1}{64}\)
\(\frac{2}{2003}< \frac{5}{2003}\)
\(\Rightarrow\text{ }A< B\)
A=2/3.(3/60.63+3/63.66+.....+3/117.120+3/120.123)
A=2/3.(1/60-1/63+1/63-1/66+...+1/117-1/120+1/20-1/123)
A=2/3.(1/60-1/123)
a) A = \(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
b) B = \(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
tính A và B
So sánh \(A=\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2006}\)và \(B=\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2006}\)
so sánh
A=\(\frac{2}{60.63}+\frac{2}{63.66}+......+\frac{2}{117.120}+\frac{2}{2017}\)
B=\(\frac{5}{40.44}+\frac{5}{44.48}+........+\frac{5}{76.80}+\frac{5}{2017}\)
A= 2x3/3x60x63+2x3/3x63x66+...+2x3/3x117x120+2/2007
=2/3(1/60-1/63+1/63-1/66+...+1/117-1/120)+2/2007
=2/3(1/60-1/120)+2/2007=1/180+2/2007
Lam tuong tu nhu vay vs B. B nhan ca tu va mau cho 4 sua do thu 5/4 ra ngoai. Cuoi cung tinh dk A<B
So sánh
C=\(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2006}\)
Và D=\(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2006}\)
C=1/2.(3/60.63+....+3/117.120)+1/1003
C=1/2.(1/60-1/63+....+1/117-1/120)+1/1003
....còn lại tự làm nha, bài còn lại cũng tương tự