\(50-\frac{50}{51}-\frac{51}{52}-\frac{52}{53}-\frac{53}{54}-.................-\frac{99}{100}\)
So sánh:
\(A=1.3.5.7...97.99\)
\(B=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.\frac{54}{2}...\frac{99}{2}.\frac{100}{2}\)
bang nhau
Giai:
A=1.3.5.7...97.99=\(\frac{\left(1.3.5...97.99\right).\left(2.4.6...100\right)}{2.4.6...100}\)
=\(\frac{1.2.3.4...99.100}{\left(1.2\right).\left(2.2\right)...\left(2.50\right)}\)
=\(\frac{\left(1.2.3...50\right).\left(51.52...99.100\right)}{\left(1.2.3...49.50\right).2^{50}}\)
=\(\frac{51.52...99.100}{2.2...2.2}\)
=\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
mà B=\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
Nên A=B
Vậy A=B
\(1.3.5.7...97.99=\frac{100!}{2.4.6.8...100}\)
\(=\frac{1.2.3.4...100}{1.2.2.2.3.2...50.2}\)
\(=\frac{51.52.53...100}{2}\)
Vậy \(A=B\)
\(A=1.3.5.....97.99\)
\(=\frac{1.2.3.4......98.99.100}{2.4.6.8......96.98.100}\)
\(=\frac{1.2.3.4....98.99.100}{2.1.2.2.2.3........49.2.50.2}\)
\(=\frac{\left(1.2.3.4......50\right)51.52....98.99.100}{2^{50}\left(1.2.3.......50\right)}\)
\(=\frac{51.52.53.....99.100}{2^{50}}\)
\(=\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot.......\cdot\frac{99}{2}\cdot\frac{100}{2}=B\)
Vậy \(A=B\)
Chứng minh rằng :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\)
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
Tính:\(\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\right):\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\right)\)
Số chia rút gọn thành 1/51+1/52+...+1/99+1/100
=> biểu thức bằng 1
so sánh:
\(c=1.3.5.7.....99\)và\(D=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.\frac{54}{2}....\frac{100}{2}\)
\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+.......+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}>\frac{1}{2}\)
ta có 1/51>1/100
1/52>1/100
..................
1/100=1/100
\(\Rightarrow\)S=1/51+1/52+...+1/100>(1/100+1/100+...+1/100)=1/100.50=1/2
\(\Rightarrow\)S>\(\frac{1}{2}\)
cái chỗ 1/100+1/100+...+1/100 có 50 số bạn nhá
chúc bạn học tốt~
Chứng minh :
\(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+....+\frac{1}{100}< 1\)
Vì mọi phân số của tổng đều nhỏ hơn 1 nên tổng đó nhỏ hơn 1.
k nha
So sánh: 1.3.5....99 với \(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}\)
C= 1.3.5.7..99 với D= \(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
Ta có :
\(C=1.3.5.7...99\Rightarrow C=\frac{1.3.5.7..99}{2.4.6.8..98}\Rightarrow C=\frac{1.3.5.7..9}{\left(2.2...2\right)\left(1.2.3..50\right)}\)( có 50 chữ số 2 )
\(\Rightarrow C=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
\(\Rightarrow C=D\)
so sánh C=\(\frac{51}{2}.\frac{52}{2}\frac{53}{2}........\frac{100}{2}\)D=1.3.5.7....99
xử lí C ta có C=51.52.53.....100/250
ta nhân cả tử và mẫu của C với 1.2.3.........50 thì dc
(1.2.3.4.5.6.........................50).(51.52..............100)
(1.2.3.4...............................50) (2.2...................2) có 50 thừa số 2
tử giữ nguyên xét mẫu ta có (1.2........50).(2.2.......2.2)= (1.2)(2.2)......(50.2)=2.4.6.8......100 vậy triệt tiêu tử cho mẫu thì ta dc c=1.3....97.99
tức C=D