Chọn đáp án đúng:
Viết 3m 45cm thành hỗn số:
A. \(3\frac{45}{10}\) m B. \(3\frac{45}{100}\) cm C. \(3\frac{45}{100}\) m
cho \(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\) và \(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
Tính tỉ số M với N
Tính:\(B=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-.....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}}\)
\(B=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
\(=\frac{\left(1-\frac{1}{9}\right)+\left(1-\frac{2}{10}\right)+....+\left(1-\frac{92}{100}\right)}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)(có 92 số 1)
\(=\frac{\frac{8}{9}+\frac{8}{10}+....+\frac{8}{100}}{\frac{1}{5}\left(\frac{1}{9}+\frac{1}{10}+....+\frac{1}{100}\right)}=\frac{8\left(\frac{1}{9}+\frac{1}{10}+....+\frac{1}{100}\right)}{\frac{1}{5}\left(\frac{1}{9}+\frac{1}{10}+....+\frac{1}{100}\right)}\)
\(=8:\frac{1}{5}=40\)
\(B\)\(=\)\(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}....+\frac{1}{500}}\)
Tham khảo bài làm bn Đàm đi
Hok tốt
Cho \(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+.......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}....+\frac{1}{100}}\)
\(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+.....+\frac{1}{495}+\frac{1}{500}}\)
Tính M; N
Cho M = \(\frac{\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)}{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\);
N = \(\frac{\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\right)}{\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}\right)}\)
Tìm tỉ số phần trăm của M và N
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 = \(\frac{100}{40}\%=250\%\)
\(M=\frac{1+(\frac{1}{99}+1)+(\frac{2}{98}+1)+(\frac{3}{97}+1)+...+(\frac{98}{2}+1)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(M=\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(M=\frac{100\cdot(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2})}{(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})}=100\)
\(N=\frac{(1-\frac{1}{9})+(1-\frac{2}{10})+(1-\frac{3}{11})+...+(1-\frac{92}{100})}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
\(N=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}=\frac{8(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100})}{\frac{1}{5}(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100})}=40\)
\(M:N=\frac{100}{40}=250\%\)
M.n giúp mình bài này vs!
M=\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
N=\(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
Tìm tỉ số phần trăm của M và N
Lm nhanh nhé!
mk nghĩ là nguyễn việt hoàng làm sai rồi!
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!
\(M=\frac{\frac{1}{99}+\frac{2}{98}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\) và \(N=\frac{92-\frac{1}{9}-\frac{2}{10}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+...+\frac{1}{500}}\)
Hãy tìm tỉ số giữa M và N
cho M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\) cho N =\(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
tính M và N rồi tính tỉ số phầm trăm của M và N
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.
ết: large frac{1}{3x10}+frac{1}{10x17}+frac{1}{17x24}+...+frac{1}{38x45}=frac{6}{x}(Chỉ được chọn 1 đáp án)A.45B.135C.\large \frac{135}{7}D.\large \frac{14}{45}
\(\dfrac{1}{3\cdot10}+\dfrac{1}{10\cdot17}+...+\dfrac{1}{38\cdot45}=\dfrac{6}{x}\)
=>\(\dfrac{1}{7}\left(\dfrac{7}{3\cdot10}+\dfrac{7}{10\cdot17}+...+\dfrac{7}{38\cdot45}\right)=\dfrac{6}{x}\)
=>\(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{38}-\dfrac{1}{45}=\dfrac{42}{x}\)
=>42/45=42/x
=>x=45
=>Chọn A
A=\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+....+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}}\)
B=\(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}}\)