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
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+)Đặt A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\left(1+1+1+...+1\right)\) (99 chữ số 1)
A= \(\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
A= \(\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+1\)
A= \(100.\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
⇒ M= \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}\)
M= \(\dfrac{100.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
M= 100 (1)
+) Đặt B= \(92-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\)
B= \(\left(1+1+1+...+1\right)-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\) ( 92 chữ số 1)
B= \(\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)
B= \(\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)
B= \(8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)
⇒ N= \(\dfrac{8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)
N= 8 (2)
Từ (1) và (2)⇒ \(\dfrac{M}{N}\) = \(\dfrac{100}{8}\)= \(\dfrac{25}{2}\)
Vậy \(\dfrac{M}{N}=\dfrac{25}{2}\)
Ta có :
M = \(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{91}+1\right)+...+\left(\frac{98}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(100\)
N = \(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{\left(1-\frac{1}{9}\right)+\left(1-\frac{2}{10}\right)+\left(1-\frac{3}{11}\right)+...+\left(1-\frac{92}{100}\right)}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{8.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{5}.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}\)
N = \(40\)
\(\Rightarrow\)M : N = \(\frac{100}{40}\%=250\%\)
Đặt: \(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(=\frac{1-\left[\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right]}{1-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]}\)
\(=\frac{1-\frac{99}{1}}{1-\frac{1}{100}}\)
\(M=\frac{-98}{99}\)
Đặt \(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
\(=\frac{92+\left[\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\right]}{1-\left[\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\right]}\)
\(=\frac{92+\frac{92}{100}}{1-\frac{1}{500}}\)
\(=\frac{92+\frac{92}{100}}{\frac{499}{500}}\)
Tự làm tiếp đi!
M : N = \(\frac{100}{40}\) = 250 %. Tính dễ nhưng mình ngại viết lắm !
tử của M=[(1/99)+(99/1)]x99/2=480298/99 tương tự mẫuM=1173/50
tử N=[22513/450] mẫu N=9919/1800
sao bấm ra số ảo quá. tính tỉ số tự tính đi.
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%