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NV
15 tháng 5 2019

Câu 2:

\(20A=\frac{20}{1.21}+\frac{20}{2.22}+\frac{20}{3.23}+...+\frac{20}{80.100}\)

\(20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+\frac{1}{3}-\frac{1}{23}+...+\frac{1}{80}-\frac{1}{100}\)

\(20A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{80}-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)

\(20A=1+\frac{1}{2}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\) (1)

Lại có:

\(B=\frac{1}{1.81}+\frac{1}{2.82}+...+\frac{1}{20.100}\)

\(\Rightarrow80B=\frac{80}{1.81}+\frac{80}{2.82}+...+\frac{80}{20.100}\)

\(80B=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+...+\frac{1}{20}-\frac{1}{100}\)

\(80B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)(2)

Từ (1) và (2) suy ra \(20A=80B\)

\(\Rightarrow\frac{A}{B}=\frac{80}{20}=4\)

NV
15 tháng 5 2019

Câu 1:

\(\frac{x}{16}-\frac{1}{y}=\frac{1}{32}\)

\(\Leftrightarrow\frac{xy-16}{16y}=\frac{1}{32}\)

\(\Leftrightarrow\frac{xy-16}{y}=\frac{1}{2}\)

\(\Leftrightarrow2xy-32=y\)

\(\Leftrightarrow\left(2x-1\right).y=32\)

Tới đây ta nhận xét do \(2x-1\) luôn lẻ với mọi x nguyên nên \(2x-1\) là ước lẻ của 32

\(\Rightarrow2x-1=\left\{1;-1\right\}\)

Vậy: \(\left\{{}\begin{matrix}2x-1=1\\y=32\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=32\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2x-1=-1\\y=-32\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-32\end{matrix}\right.\)

Có 2 cặp số nguyên thỏa mãn là \(\left(x;y\right)=\left(1;32\right);\left(0;-32\right)\)

24 tháng 4 2019

20A=20/1.21+20/2.22+...+20/80.100

=1-1/21+1/2-1/22+...+1/80-1/100

=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)

80B=80/1.81+80/2.82+...+8/20.100

=1-1/81+1/2-1/82+...+1/20-1/100

=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)

=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)

=>20A=80B

=>A=4B

17 tháng 3 2020

ta có: \(A=\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+...+\frac{1}{80.100}\)

\(20A=\frac{20}{1.21}+\frac{20}{2.22}+\frac{20}{2.23}+...+\frac{20}{80.100}\)

\(20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+\frac{1}{3}-\frac{1}{23}+...+\frac{1}{80}-\frac{1}{100}\)

\(20A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{80}-\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{100}\right)\)

\(20A=1+\frac{1}{2}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+\frac{1}{83}+...+\frac{1}{100}\right)\)

lại có: \(B=\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+...+\frac{1}{20.100}\)

\(80B=\frac{80}{1.81}+\frac{80}{2.82}+\frac{80}{3.83}+...+\frac{80}{20.100}\)

\(80B=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+\frac{1}{3}-\frac{1}{83}+...+\frac{1}{20}-\frac{1}{100}\)

\(80B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+\frac{1}{83}+...+\frac{1}{100}\right)\)

Vậy 20A = 80B

=> \(\frac{A}{B}=\frac{80}{20}=4\)

17 tháng 3 2020

\(A=\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+...+\frac{1}{80.100}\)

\(20A=\frac{20}{1.21}+\frac{20}{2.22}+\frac{20}{3.23}+...+\frac{20}{80.100}\)

\(20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+\frac{1}{3}-\frac{1}{23}+...+\frac{1}{80}-\frac{1}{100}\)

\(20A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{80}-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)

\(20A=1+\frac{1}{2}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)(1)

Lại có : 

\(B=\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+...+\frac{1}{20.100}\)

\(\Rightarrow80B=\frac{80}{1.81}+\frac{80}{2.82}+...+\frac{80}{20.100}\)

\(80B=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+...+\frac{1}{20}-\frac{1}{100}\)

\(80B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)(2)

Từ (1) và (2) , suy ra : \(20A=80B\)

\(\Rightarrow\frac{A}{B}=\frac{80}{20}=4\)

Y
18 tháng 5 2019

3. + \(20A=\frac{21-1}{1\cdot21}+\frac{22-2}{2\cdot22}+...+\frac{100-80}{80\cdot100}\)

\(\Rightarrow20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+...+\frac{1}{80}-\frac{1}{100}\)

\(\Rightarrow20A=\left(1+\frac{1}{2}+...+\frac{1}{80}\right)-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)

\(\Rightarrow A=\frac{1}{20}\left[\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\right]\)

+ \(80B=\frac{81-1}{1\cdot81}+\frac{82-2}{2\cdot82}+...+\frac{100-2}{20\cdot100}\)

\(=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+...+\frac{1}{20}-\frac{1}{100}\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)

\(\Rightarrow B=\frac{1}{80}\left[\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\right]\)

Do đó : \(\frac{A}{B}=\frac{\frac{1}{20}}{\frac{1}{80}}=4\)

4. + \(A=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}=\frac{100}{2}=50\)

18 tháng 4 2022

Giúp mình đi các bạn ơi

22 tháng 4 2023

A=20/1.21+20/2.22+...+20/80.100

=1-1/21+1/2-1/22+...+1/80-1/100

=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)

80B=80/1.81+80/2.82+...+8/20.100

=1-1/81+1/2-1/82+...+1/20-1/100

=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)

=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)

=>20A=80B

=>A=4B

29 tháng 4 2023

A=20/1.21+20/2.22+...+20/80.100

=1-1/21+1/2-1/22+...+1/80-1/100

=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)

80B=80/1.81+80/2.82+...+8/20.100

=1-1/81+1/2-1/82+...+1/20-1/100

=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)

=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)

=>20A=80B

=>A=4B

13 tháng 4 2023

20a = 20/1.21 + 20/2.22+ ... + 20/80.100

= 1-1/21 + 1/2 - 1/22 +...+ 1/80 - 1/100

= 1  + 1/2 + 1/3 +... + 1/19 + 1/20 - 1/81 - 1/82 -.... - 1/100

80b = 80/1.81 + 80/2.82 + 80/3.83 +... + 80/20.100

= 1 - 1/81+ 1/2 - 1/83 +...+ 1/20 - 1/100

=> 20a = 80b

=> a/b = 4