(1+1/1.3).(1+1/2.4)(1+1/3.5)....(1+1/99.101)
(1+1/1.3).(1+1/2.4).(1+1/3.5).........(1+1/99.101)
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(\frac{1}{3.5.}\right).....\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{10000}{9999}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(=\frac{2^2.3^2.4^2.5^2.....98^2.99^2.100^2}{1.2.3^2.4^2.5^2......99^2.100.101}\)
\(=\frac{2.100}{1.101}\)
\(=\frac{200}{101}\)
(1+1/1.3)(1+1/2.4)(1+1/3.5)...(1/1+1/99.101)
A=(1+1/1.3).(1+1/2.4).(1+1/3.5).....(1+1/99.101)
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....+\left(1+\frac{1}{99.101}\right)\)
\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(=\frac{2.3.4.....100}{1.2.3.....99}.\frac{2.3.4.....100}{3.4.5.....101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
Tính H = (1 + 1/1.3)(1 + 1/2.4)(1 + 1/3.5)...(1+1/99.101)
V=(1+1/1.3).(1+1/2.4).(1+1/3.5)....(1+1/99.101)
Tính nhanh
Tính
D=(1+1/1.3)(1+1/2.4)(1+1/3.5).............(1+1/99.101)
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}....\frac{10000}{99.101}\)
\(=\frac{2.2.3.3...100.100}{1.3.2.4...99.101}\)
\(=\frac{\left(2.3.4...100\right)\left(2.3.4...100\right)}{\left(1.2...99\right)\left(3.4.5...101\right)}\)
\(=\frac{100.2}{101}=\frac{200}{101}\)
\(D=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(D=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{10000}{99.101}\)
\(D=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{100^2}{99.101}\)
\(D=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4.5...101}=100.\frac{2}{101}=\frac{200}{101}\)
Vậy \(D=\frac{200}{101}\)
Tính
A=(1-1/2).(1-1/3).(1-1/4).....(1-1/100)
B= (1+1/1.3).(1+1/2.4).(1+1/3.5).....(1+1/99.101)
Tính nhanh:
M = ( 1 + 1/1.3 ) . ( 1 + 1/2.4 ) . ( 1 + 1/3.5 ) .... ( 1 + 1/99.101 )
N = 1.3.5 + 2.6.10 + 4.12.20 + 7.21.35 / 1.3.5 + 2.10.14 + 4.20.28 + 7.35.49
Xét số hạng tổng quát:
1 + 1/[k.(k + 2)] = [k.(k + 2) + 1]/[k.(k + 2)] = (k + 1)²/[k.(k + 1)], với k nguyên dương.
Cho k chạy từ 1 đến 99, ta có:
• 1 + 1/1.3 = 2²/(1.3).
• 1 + 1/2.4 = 3²/(2.4).
• 1 + 1/3.5 = 4²/(3.5).
.......................
• 1 + 1/97.99 = 98²/(97.99).
• 1 + 1/98.100 = 99²/(98.100).
• 1 + 1/99.101 = 100²/(99.101).
Nhân vế với vế các đẳng thức trên, ta được:
(1 + 1/1.3).(1 + 1/2.4)(1 + 1/3.5)....(1 + 1/99.101)
= [2².3².....100²]/[1.2.3².4²......99².100...
= (2².100²)/(2.100.101)
= 2.100/101
= 200/101.
còn N thì chịu
M=(4/1.3.9/2.4.16/3.5...10000/99.101
M=2.2/1.3.3.3/2.4.4.4/3.5...100.100/99.101
M=2.3.4.5...100/1.2.3...99.3.4.5...100/2.3.4.5...101
M=100.2/101=200/101
Cau N sai de rui ban a, o mau so phai la 1.5.7+2.10.14+4.20.28+7.35.49 moi lam dc.
mấy bn giải khó hiểu wa .....................
Bài 3 : (1+1/1.3)(1+1/2.4)(1+1/3.5)...(1+1/99.101)
bài 4:
(x-3)(x+5)/(x-2)^2 <0
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{99.101}\right)\)
\(=\frac{2.2}{1.3}\frac{3.3}{2.4}.....\frac{100.100}{99.101}\)
\(=\frac{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}\)
\(=\frac{100.2}{101}=\frac{200}{101}\)
\(\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right)^2}< 0\)
\(\Rightarrow\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right).\left(x-2\right)}< 0\)
=> ( x - 3 ) . ( x - 5 ) và ( x - 2 ) . ( x - 2 ) trái dấu
Mà ( x - 2 )2 = ( x - 2 ) . ( x - 2 ) ≥ 0 ∀ x
\(\Rightarrow\hept{\begin{cases}\left(x−3\right).\left(x+5\right)< 0\\\left(x−2\right).\left(x−2\right)>0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< −5;−5< x< 3\\x>2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< −5\\2< x< 3\end{cases}}\)