\(G=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{99}{2}.\frac{100}{2}-1.3.5.7.9...97.99\)
tính
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 \(A=B\)
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
thực hiện tính và so sánh
C=1.3.5.7.........99 với D=\(\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot\cdot\cdot\cdot\cdot\cdot\frac{100}{2}\)
So sánh: C=1.3.5.7...99 và \(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
-->C=\(\frac{1.2.3.4...99.100}{2.4.6....100}\)-->C=\(\frac{1.2.3...99.100}{\left(2.2....2\right)\left(1.2.3.4.5....50\right)}\)[50 chữ số 2]
-->\(C=\frac{51}{2}.\left(\frac{52}{2}\right)....\left(\frac{100}{2}\right)\)=D vậy C=D
________________________________________________________
LI-KE CHO MK NHÉ BN
so sánh: C=1.3.5.7...99 với D= \(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
-->C=\(\frac{1.2.3.4...99.100}{2.4.6....100}\)-->C=\(\frac{1.2.3...99.100}{\left(2.2....2\right)\left(1.2.3.4.5....50\right)}\)[50 chữ số 2]
-->\(C=\frac{51}{2}.\left(\frac{52}{2}\right)....\left(\frac{100}{2}\right)\)=D vậy C=D
________________________________________________________
LI-KE CHO MK NHÉ BN
so sánh: C=1.3.5.7...99 với D= 512 .522 .532 ...1002
Tính: \(B=\frac{100^2+1^2}{100\cdot1}+\frac{99^2+2^2}{99\cdot2}+\frac{98^2+3^2}{98\cdot3}+...+\frac{52^2+49^2}{52\cdot49}+\frac{51^2+50^2}{51\cdot50}\)
so sánh:
\(C=1.3.5.7...99\) và\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}\)
C=1.3.5.7...99
=>2.4.6...100.C=1.2.3...100
=>C = (1.2.3....100) / (2.4.6...100)= (1.2.3...50).(51.52...100) / [(2.1)(2.2).(2.3)...(2.50)]
C=(1.2.3...50).(51.52...100) /[2^50.(1.2.3...50)] =(51.52...100)/2^50 =51/2.52/2.53/2...100/2 =D
VAy C=D