`M=(√x+1)/(√x-1)` đk: `x≥0` và`x`$\neq$`1`
`M²≤1`
`⇒((√x+1)/(√x-1))²≤1`
`⇔((√x+1)²)/((√x-1)²)≤1`
`⇔(x+2√x+1)/(x-2√x+1)≤(x-2√x+1)/(x-2√x+1)`
`⇔ (x+2√x+1)/(x-2√x+1)-(x-2√x+1)/(x-2√x+1)≤0`
`⇔(x+2√x+1-x+2√x-1)/(x+2√x+1)≤0`
`⇔(4√x)/(x+2√x+1)≤0`
`⇔4√x≤0`
`⇔√x≤0`
`⇔x≤0`
Mà `x≥0`
Vậy để`M²≤1` thì `x=0`
#NoCopy