Đáp án:Nhìn mấy bài này khá căn bản thôi chỉ cần quy đồng là xong nhé!
Giải thích các bước giải:
`A=((sqrtx-2)/(x-1)-(sqrtx-2)/(x+2sqrtx+1)).(x^2-2x+1)/2`
Điều kiện:`x>=0,x ne 1`
`A=((sqrtx-2)/(x-1)-(sqrtx-2)/(x+2sqrtx+1)).(x-1)^2/2`
`=(((sqrtx-2)(sqrtx+1))/((x-1)(sqrtx+1))-((sqrtx+2)(sqrtx-1))/((x-1)(sqrtx+1))).(x-1)^2/2`
`=(((sqrtx-2)(sqrtx+1)-(sqrtx+2)(sqrtx-1))/((x-1)(sqrtx+1))).(x-1)^2/2`
`=(((x-sqrtx-2-x-sqrtx+2))/((x-1)(sqrtx+1))).(x-1)^2/2`
`=(-2sqrtx(x-1)^2)/(2(x-1)(sqrtx+1))`
`=sqrtx(1-sqrtx)`
Vậy `A=sqrtx(1-sqrtx)`.
`b)x<1`
`=>sqrtx<1`
`=>1-sqrtx>0`
`<=>sqrtx(1-sqrtx)>0`
`=>A>0AA0<x<1`
Vậy với `0<x<1` thì `A>0`.
`c)A=3+2sqrt2=2+2sqrt2+1=(sqrt2+1)^2`
`=>A=(sqrt2+1)[1-(sqrt2+1)]`
`=(sqrt2+1).(-sqrt2)`
`=-2-sqrt2`
Vậy `x=3+2sqrt2` thì `A=-2-sqrt2`.
`d)A=sqrtx(1-sqrtx)`
`=-sqrtx(sqrtx-1)`
`=-(sqrtx-1/2+1/2)(sqrt-1/2-1/2)`
`=-[(sqrtx-1/2)^2-1/4]`
`=-(sqrtx-1/2)^2+1/4<=1/4`
Dấu "=" xảy ra khi `sqrtx=1/2<=>x=1/4`
Vậy `Max_A=1/4<=>x=1/4`.
