Giải thích các bước giải:
b, $T$a có :
$\dfrac{x}{x+y+z} > \dfrac{x}{x+y+z+t}$
$\dfrac{y}{x+y+t} > \dfrac{y}{x+y+z+t}$
$\dfrac{z}{y+z+t} > \dfrac{z}{x+y+z+t}$
$\dfrac{t}{z+t+x} > \dfrac{t}{x+y+z+t}$
⇒ $\dfrac{x}{x+y+z} + \dfrac{y}{x+y+t} + \dfrac{z}{y+z+t} + \dfrac{t}{z+t+x}$
> $\dfrac{x}{x+y+z+t} + \dfrac{y}{x+y+z+t} + \dfrac{z}{x+y+z+t} + \dfrac{t}{x+y+z+t}$
= $\dfrac{x+y+z+t}{x+y+z+t}$
= $1$
⇒ $M > 1 (1)$
$T$a có :
$\dfrac{x}{x+y} > \dfrac{x}{x+y+z}$
$\dfrac{y}{x+y} > \dfrac{y}{x+y+t}$
$\dfrac{z}{z+t} > \dfrac{z}{y+z+t}$
$\dfrac{t}{z+t} > \dfrac{t}{z+t+x}$
⇒ $\dfrac{x}{x+y+z} + \dfrac{y}{x+y+t} + \dfrac{z}{y+z+t} + \dfrac{t}{z+t+x}$
< $\dfrac{x}{x+y} + \dfrac{y}{x+y} + \dfrac{z}{z+t} + \dfrac{t}{z+t}$
= $(\dfrac{x}{x+y} + \dfrac{y}{x+y}) + (\dfrac{z}{z+t} + \dfrac{t}{z+t}$)
= $\dfrac{x+y}{x+y} + \dfrac{z+t}{z+t}$
= $1 + 1$
= $2$
⇒ $M < 2 (2)$
$T$ừ $(1) và (2)$
⇒ $M ∉ N (1 < M < 2)$
$C$húc bạn học tốt !
