Bài 5:
Ta có: \(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\cdots+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
=>\(4S=1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\cdots+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\)
=>\(4S+S=1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\cdots+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}+\frac{1}{2^2}-\frac{1}{2^4}+\cdots+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
=>\(5S=1-\frac{1}{2^{2004}}<1\)
=>\(S<\frac15\)
=>S<0,2
Bài 3: Sửa đề: x,y nguyên
c: x+y+9=xy-7
=>xy-7-x-y-9=0
=>xy-x-y-16=0
=>x(y-1)-y+1-17=0
=>(x-1)(y-1)=17
=>(x-1;y-1)∈{(1;17);(17;1);(-1;-17);(-17;-1)}
=>(x;y)∈{(2;18);(18;2);(0;-16);(-16;0)}
b:
Ta có: \(x^3y=xy^3+1997\)
=>\(x^3y-xy^3=1997\)
=>\(xy\left(x^2-y^2\right)=1997\)
=>xy(x-y)(x+y)=1997
Đặt A=xy(x-y)(x+y)
TH1: x chẵn; y chẵn
=>xy chẵn
=>xy(x-y)(x+y)⋮2
=>A⋮2(1)
TH2: x chẵn, y lẻ
=>xy chẵn
=>xy(x-y)(x+y)⋮2
=>A⋮2(2)
TH3: x lẻ; y chẵn
=>xy chẵn
=>A=xy(x-y)(x+y)⋮2(3)
TH4: x lẻ; y lẻ
=>x+y chẵn
=>(x+y)(x-y)xy⋮2
=>A⋮2(4)
Từ (1),(2),(3),(4) suy ra A⋮2
mà A=1997
và 1997 không chia hết cho 2
nên (x;y)∈∅
