Đáp án+Giải thích các bước giải:
Bài 1:
1. ĐKXĐ: `x>0;x\ne1`
`P=[((\sqrtx+2)(\sqrtx+1))/((\sqrtx+2)(\sqrtx -1)) - (\sqrtx(\sqrtx+1))/((\sqrtx+1)(\sqrtx-1))]: (\sqrtx-1+\sqrtx+1)/((\sqrtx+1)(\sqrtx-1))`
`P= [(\sqrtx+1)/(\sqrtx -1) -(\sqrtx)/(\sqrtx-1)] : (2\sqrtx)/((\sqrtx+1)(\sqrtx-1))`
`P=1/(\sqrtx-1) . ((\sqrtx+1)(\sqrtx-1))/(2\sqrtx)`
`P=(\sqrtx+1)/(2\sqrtx)`
Vậy `P=(\sqrtx +1)/(2\sqrtx)` với `x>0;x\ne1.`
2. Có `x=3-2\sqrt2=(\sqrt2-1)^2 in` ĐKXĐ
`=>\sqrtx=\sqrt((\sqrt2-1)^2)=|\sqrtx2-1|=\sqrt2-1 ( vì \sqrt2-1>0)`
Thay `\sqrtx=\sqrt2-1` vào P, ta có:
`P=(\sqrt2-1+1)/(2(\sqrt2-1))`
`P=(\sqrt2)/(2\sqrt2-2) =1/(2-\sqrt2)`
`P=(2+\sqrt2)/(4-2)`
`P=(2+\sqrt2)/2`
Vậy `P=(2+\sqrt2)/2` khi `x=3-2\sqrt2.`
