Question

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:\(\frac{A}{B}\)

Answer 1

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 \(20A=80B\)

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

Answer 2

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 \(\left(x;y\right)=\left(1;32\right);\left(0;-32\right)\)