Cho \(A=\dfrac{1}{1.21}+\dfrac{1}{2.22}+\dfrac{1}{3.23}+...+\dfrac{1}{80.100}\);
\(B=\dfrac{1}{1.81}+\dfrac{1}{2.82}+\dfrac{1}{3.83}+...+\dfrac{1}{20.100}\).
Tính \(\dfrac{A}{B}\).
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18 tháng 4 2022 lúc 14:54
Câu 24 : Cho A = 1.21 + 1/2.22 + 1/3.23 + ... + 1/80.100 ; B = 1/1.81 + 1/2.82 + 1/3.83 + ... + 1/20.100 . Tính A/B
tính a/b biết a=1/1.21 + 1/2.22 + 1/3.23 + ... +1/80.100; b=1/1.81 + 1/2.82 + 1/3.82 + ... + 1/20.100
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
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Cho \(A=\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+......+\frac{1}{80.100}\)
\(B=\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+....+\frac{1}{20.100}\). Tính \(\frac{A}{B}\)
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
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18 tháng 4 2022 lúc 14:51
Câu 24 : Cho A = 1/1.21 + 1/2.22 + 1/3.23 + ...+ 1/80.10
cho a = 1/1.21 + 1/2.22+...+1/80.100 b= 1/1.81 + 1/2.82+...+1/20.100 tính a/b
Cho A=1/1.21+1/1.22+1/3.23+...+1/80.100
B=1/1.81+1/2.82+1/3.83+...+1/20.100
Tính A/B
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
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Tính \(\frac{A}{B}\)
A=\(\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+.....+\frac{1}{80.100}\)
B=\(\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+.....+\frac{1}{20.100}\)
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
=> \(\frac{A}{B}=\frac{80}{20}=4\)
\(A=\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+...+\frac{1}{80.100}\)
\(20A=\frac{20}{1.21}+\frac{20}{2.22}+\frac{20}{3.23}+...+\frac{20}{80.100}\)
\(20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+\frac{1}{3}-\frac{1}{23}+...+\frac{1}{80}-\frac{1}{100}\)
\(20A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{80}-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)
\(20A=1+\frac{1}{2}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)(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 : \(20A=80B\)
\(\Rightarrow\frac{A}{B}=\frac{80}{20}=4\)
A=\(\frac{1}{1.21}\)+\(\frac{1}{2.22}\)+\(\frac{1}{3.23}\)+...+\(\frac{1}{80.100}\)và B=\(\frac{1}{1.81}\)+\(\frac{1}{2.82}\)+\(\frac{1}{3.83}\)+...+\(\frac{1}{20.100}\).Tính \(\frac{A}{B}\)
1) Tìm các cặp số nguyên (x,y) sao cho \(\frac{x}{16}-\frac{1}{y}=\frac{1}{32}\)
2) Cho A=\(\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+...+\frac{1}{80.100}\)
B=\(\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+...+\frac{1}{20.100}\) Tính:\(\frac{A}{B}\)
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 \(20A=80B\)
\(\Rightarrow\frac{A}{B}=\frac{80}{20}=4\)
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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 \(\left(x;y\right)=\left(1;32\right);\left(0;-32\right)\)
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