Question

                So sánh a và b ,biết:

                                     \(\frac{2+4+6+8+...+2a}{a}< \frac{2+4+6+8+...+2b}{b}\)

Answer 1

Đặt \(A=\frac{2+4+...+2a}{a};B=\frac{2+4+...+2b}{b}\)

Ta có: \(A=\frac{2+4+...+2a}{a}=\frac{2\left(1+2+...+a\right)}{a}=\frac{\frac{2a\left(a+1\right)}{2}}{a}=\frac{a\left(a+1\right)}{a}=a+1\)

\(B=\frac{2+4+...2b}{b}=\frac{2\left(1+2+...+b\right)}{b}=\frac{\frac{2b\left(b+1\right)}{2}}{b}=\frac{b\left(b+1\right)}{b}=b+1\)

Vì A < B => a+1 < b + 1 

Xét a,b thuộc Z+ => a < b

Xét a,b thuộc Z- => a > b

Answer 2

2+4+6+8+..............2a=2(1+2+3+.............+a)=2.(a+1).a

=>2+4+6+8+..............2a/a=2.(a+1)

2+4+6+8+..............2b=2(1+2+3+.............+b)=2.(b+1).b

=>2+4+6+8+..............2b/b=2.(b+1)

Vì 2.(a+1)<2.(b+1)

=>a+1<b+1

=>a<b

Vậy a<b