cho T = 1/2 + 1/3 + 1/4 +....+ 1/99 + 1/100 và M = 1/99 + 2/98 + 3/97 + ...+ 97/3 + 98/2 +99/1
hãy tìm tỉ số T/M
M=
1/99 + 2/98 + 3/97 + ... + 99/1
1/99 + 2/98 + 3/97 + ... + 99/1
N=
92 - 1/9 - 2/10 - 3/11 - ... - 92/100
1/45 + 1/50 + 1/55 + ...+ 1/500
Tìm tỉ số phần trăm của M và N
M=1
N = 1-1/9 + 1-2/10 + 1-3/11 +...+ 1-92/100/1/45+1/50+1/55+...+1 /500
= 8/9+8/10+8/11+8/12+...+8/100 / 1/5.9+1/5.10+1/5.11+...+1/ 5.100
= 8 .(1/9+1/10+1/11+...+1/100) / 5 .(1/9+1/10+1/11+...+1/100)
= 8/5
vậy tỉ số phần trăm của M và N là: 1:8/5= 62,5%
Tính M/N biết M = 99/1 + 98/2 + 97/3 + ... + 2/98 + 1/99 và N = 1/2 + 1/3 + 1/4 +...+1/99+1/100
\(M=\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
cộng vào mỗi phân số trong 98 phân số sau,trừ phân số cuối đi 98 , ta được :
\(M=1+\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{2}{98}+1\right)+\left(\frac{1}{99}+1\right)\)
\(M=\frac{100}{100}+\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
chuyển phân số \(\frac{100}{100}\)ra sau , ta được :
\(M=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}+\frac{100}{100}\)
\(M=100.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{100.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}}=100\)
Giải giúp tớ bài này với
Cho A = 1/99+2/98+3/97+4/96+....+98/2+99/1 và B = 1/2+1/3+1/4+...+1/100
Tính tỉ số A và B
\(A=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+\frac{4}{96}+...+\frac{98}{2}+\frac{99}{1}\)
\(A=1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{97}+1\right)+\left(\frac{4}{96}+1\right)+...+\left(\frac{98}{2}+1\right)\)
\(A=\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+\frac{100}{96}+...+\frac{100}{2}\)
\(A=100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=100\)
Tính nhanh
a, 1-2+3-4+.....+2015-2016+2017
b,1+3-5-7+9+11+....+97-98-99+100+101
c,1-2-3+4+5-6-7+....+97-98-99+100+101
d,2^100-2^99-2^98-....-2-1
Nhanh nha m dang cần gấp
tìm m =1/99+2/98+3/97+.....+99/1<đây là tử số>
1/.2+1/3+1/4......+1/100<mấu số>
Tử số: \(T=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)
\(T=\frac{1}{99}+1+\frac{2}{98}+1+\frac{3}{97}+1+...+\frac{98}{2}+1+\frac{99}{1}+1-99\)
\(T=\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}+1=100\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
Trong ngoặc chính là mẫu số nên
m=100.
99/1+98/2+97/3+................+2/98+1/99
1/2+1/3+1/4+.......................+1/99+1/100
C=(1/2+1/3+ 1/4+... +1/100): (99/1 + 98/2 + 97/3 ... +1/99)
tìm C
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
1)11-12+13-14+15-16+17-18+19-20+21-22+.........+99-100
2)2-4+6-8+......+1998-2000
3)-1+3-5+7-....+97-99
4)1+2-3-4+.........+97+98-99-100
5)1-2+3-4+.............+99-100
6)1+3-5-7+......+97-98-99+100
7)2100-299-298-..........22-2-1
8)1-4+7-10+........+307-310+313