Đáp án+Giải thích các bước giải:
`1)2\sqrtx-x(x>=0)`
`=-(x-2\sqrtx)`
`=-(x-2\sqrtx+1-1)`
`=-(\sqrtx-1)^2+1`
Vì `-(\sqrtx-1)^2<=0`
`=>-(\sqrtx-1)^2+1<=1`
Dấu "=" xảy ra khi `\sqrtx=1<=>x=1.`
Vậy `GTLN_1=1<=>x=1.`
`2)6\sqrtx-x(x>=0)`
`=-(x-6\sqrtx)`
`=-(x-6\sqrtx+9-9)`
`=-(\sqrtx-3)^2+9`
Vì `-(\sqrtx-3)^2<=0`
`=>-(\sqrtx-3)^2+9<=9`
Dấu "=" xảy ra khi `\sqrtx=3<=>x=9`.
Vậy `GTLN_2=9<=>x=9.`
`3)1/\sqrt{x^2+1}`
Vì `x^2+1>=1>0AAx`
`=>\sqrt{x^2+1}>=1>0AAx`
`=>1/\sqrt{x^2+1}<=1/1=1`
Dấu "=" xảy ra khi `x^2=0<=>x=0`.
Vậy `GTLN_3=1<=>x=0.`
`4)1/(2x-\sqrtx+3)(x>=0)`
Ta có:
`2x-\sqrtx+3`
`=2(x-\sqrtx/2)+3`
`=2(x-2*\sqrtx*1/4+1/16-1/16)+3`
`=2(\sqrtx-1/4)^2+3-1/8`
`=2(\sqrtx-1/4)^2+23/8>=23/8`
`=>1/(2x-\sqrtx+3)<=1:23/8=8/23`
Dấu "=" xảy ra khi `\sqrtx=1/4<=>x=1/16`
Vậy `GTLN_4=8/23<=>x=1/16.`
