2) \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
Vì \(1-2y\) luôn là số lẻ nên \(1-2y\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow y=\left\{0;1;-2;3\right\}\)
\(\Rightarrow x\in\left\{40;-40;8;-8\right\}\)
Vậy các cặp số x,y thỏa mãn là \(\left(0;40\right);\left(1;-40\right);\left(-2;8\right);\left(3;-8\right)\)
Ta có :
\(B=\dfrac{2000+2001}{2001+2002}=\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}\)
Mặt khác :
\(\dfrac{2000}{2001}>\dfrac{2000}{2001+2002}\)
\(\dfrac{2001}{2002}>\dfrac{2001}{2001+2002}\)
\(\Leftrightarrow A=\dfrac{2000}{2001}+\dfrac{2001}{2002}>\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}=\dfrac{2000+2001}{2001+2002}=B\)
\(\Leftrightarrow A>B\)
Ta có: B =20002001+2002 +20012001+2002
Mặt khác: 20002001 >20002001+2002
20012002 >20012001+2002
Suy ra 20002001 +20012002 >20002001+2002 +20012001+2002
hay A> B
Vậy A > B.