Question

[tran duc huy]

Gia su x = \(\dfrac{a}{m}\), y = \(\dfrac{b}{m}\) (a,b \(\in\)Z ; m>0) va x<y

Hay chung to rang z = \(\dfrac{2a+1}{2m}\) thi ta co x<z<y

Answer 1

[Ju Moon Adn]

Ta co : x<y =>\(\dfrac{a}{m}< \dfrac{b}{m}\Rightarrow a< b\)

\(x=\dfrac{a}{m}=\dfrac{2a}{2m}\)

\(y=\dfrac{b}{m}=\dfrac{2b}{2m}\)

\(z=\dfrac{2a+1}{2m}\)

do 2a < 2a+1 => \(\dfrac{2a}{2m}< \dfrac{2a+1}{2m}\)=> x<z ⁽¹⁾

a<b => a+1 \(\le\)b

\(\Rightarrow2a+2\le2b\)

\(\Rightarrow2a+1< 2b\)

\(\Rightarrow\dfrac{2a+1}{2m}< \dfrac{2b}{2m}\)

\(\Rightarrow z< y\) ⁽²⁾

^{\(Tu\left(1\right)va\left(2\right)\)}

\(\Rightarrow x< z< y\)

Answer 2

[lê thị hương giang]

Gia su x = \(\dfrac{a}{m}\), y = \(\dfrac{b}{m}\) (a,b Z ; m>0) va x<y

Hay chung to rang z = \(\dfrac{2a+1}{2m}\) thi ta co x<z<y

Giải

x = \(\dfrac{a}{m}\), y = \(\dfrac{b}{m}\)

mà x < y => a < b

=> \(x=\dfrac{2a}{2m};y=\dfrac{2b}{2m}\)

Ta có : a < b

=> a + a < a + a + 1

=> 2a < 2a + 1

=> \(\dfrac{2a}{2m}< \dfrac{2a+1}{2m}\) hay x < z (1)

Ta có : a < b

=> a + a + 1 < b + b

=> 2a+ 1 < 2b

=> \(\dfrac{2a+1}{2m}< \dfrac{2b}{2m}\) hay z < y (2)

Từ (1) và (2) => x < y <z