Đáp án+Giải thích các bước giải:
Bài 1:
Với `x>0`
Ta có:
`A=2/\sqrtx`
`B=(2\sqrtx+2)/(\sqrtx)`
`⇒B/A=(2\sqrtx+2)/\sqrtx:2/\sqrtx`
`=(2\sqrtx+2)/(\sqrtx).(\sqrtx)/2`
`=(2\sqrtx+2)/2`
`=(2(\sqrtx+1))/2`
`=\sqrtx+1`
Xét hiệu: `B/A-1`
`=\sqrtx+1-1`
`=\sqrtx`
Vì `x>0`
`⇒\sqrtx>0`
`⇒B/A-1>0`
`⇒B/A>1`
Vậy `B/A>1`
Bài 2:
`a)`
Với `x>=0;x\ne1`
Ta có:
`M=(x+2)/(x\sqrtx-1)+(\sqrtx+1)/(x+\sqrtx+1)-1/(\sqrtx-1)`
`=(x+2)/((\sqrtx-1)(x+\sqrtx+1))+(\sqrtx+1)/(x+\sqrtx+1)-1/(\sqrtx-1)`
`=(x+2+(\sqrtx+1)(\sqrtx-1)-(x+\sqrtx+1))/((\sqrtx-1)(x+\sqrtx+1))`
`=(x+2+x-1-x-\sqrtx-1)/((\sqrtx-1)(x+\sqrtx+1))`
`=((x+x-x)-\sqrtx+(2-1-1))/((\sqrtx-1)(x+\sqrtx+1))`
`=(x-\sqrtx)/((\sqrtx-1)(x+\sqrtx+1))`
`=(\sqrtx(\sqrtx-1))/((\sqrtx-1)(x+\sqrtx+1))`
`=\sqrtx/(x+\sqrtx+1)`
Vậy với `x>=0;x\ne1` thì `M=\sqrtx/(x+\sqrtx+1)`
`b)`
Xét hiệu:
`M-1/3`
`=\sqrtx/(x+\sqrtx+1)-1/3`
`=(3\sqrtx-(x+\sqrtx+1))/(3(x+\sqrtx+1))`
`=(3\sqrtx-x-\sqrtx-1)/(3(x+\sqrtx+1))`
`=(-x+2\sqrtx-1)/(3(x+\sqrtx+1))`
`=(-(x-2\sqrtx+1))/(3(x+\sqrtx+1))`
`=(-(\sqrtx-1)^2)/(3(x+\sqrtx+1))`
Vì:
`x≥0`
`⇒x+\sqrtx>=0`
`⇒x+\sqrtx+1>=1>0`
`⇒3(x+\sqrtx+1)>0` `(1)`
Vì `x>=0;x\ne1`
`⇒(\sqrtx-1)^2>0`
`⇒-(\sqrtx-1)^2<0` `(2)`
Từ `(1)` và `(2)`
`⇒(-(\sqrtx-1)^2)/(3(x+\sqrtx+1))<0`
`⇒M-1/3<0`
`⇒M<1/3`
Vậy `M<1/3`
