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a) Ta có:
(n-1)/n < n/(n+1)
vì (n-1).(n+1)=n2-1 < n2
=>
1/2 < 2/3
3/4 < 4/5
....
99/100 < 100/101
Vậy A < B
b). Ta lại có:
A.B = 1/2 . 2/3 . 3/4 . 4/5 .... . 99/100 . 100/101 = 1/100
Mà A<B => A.A<A.B=1/100
=> A2 < 1/100
=> A < 1/10<1
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
Ta có:
\(A=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Leftrightarrow3A=99.100.101\Leftrightarrow A=\frac{99.100.101}{3}=333300\)
\(B=1.2.3+2.3.4+4.5.6+...+98.99.100\)
\(\Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+4.5.6.\left(7-3\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100\)
\(\Leftrightarrow4B=98.99.100.101\Leftrightarrow B=\frac{98.99.100.101}{4}=24497550\)
câu1:-45.99
=-45.(100-1)
=-45.100-(-45).1
=-4500-(-45)
=-4455
a. Vì
1/2<2/3
3/4<4/5
.........
99/100<100/101 nên M<N
b.M.N=\(\frac{1.2.3.4......100}{2.3.4.5......101}\)=\(\frac{1}{101}\)
tính vế trên trước: 2^12 x 3^5 - 4^6 x 9^2 = 4096 x 243 - 4096 x 81 = 4096 x ( 243 - 81 ) = 663552.
tính vế dưới sau : ( 2^2 x 3 )^6 + 8^4 x 3^5 = ( 4 x 3 )^6 + 4096 x 243 = 12^6 + 995328 = 2985984 + 995328 = 3981312.
Vậy được: 663552/3981312 = 1/6.
a, 2 + 4 + 6 +...+ 2x = 210
=> 2(1 + 2 + 3 +...+ x) = 210
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 210
=> x(x + 1) = 14.15
=> x = 14
b, Ta có: \(B=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}=\frac{51.52.53....100}{2^{50}}\)
\(=\frac{\left(51.52.53....100\right)\left(1.2.3.....50\right)}{2^{50}\left(1.2.3.....50\right)}\)
\(=\frac{1.2.3.....100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)....\left(2.50\right)}\)
\(=\frac{\left(1.3.5....99\right)\left(2.4.6....100\right)}{2.4.6.....100}\)
\(=1.3.5.....99=B\)
Vậy A = B
a. Vì
1/2 < 2/3
3/4 < 4/5
..........
99/100<100/101 nên M<N
b.M.N=\(\frac{1.2.3.4.........100}{2.3.4.5.........101}=\frac{1}{101}\)