Bài `29:`

`a)`

`P=[(1/sqrtx+1/sqrty). 2/(sqrtx+sqrty)+1/x+1/y]:(sqrt(x^3)+ysqrtx+xsqrty+sqrt(y^3))/(sqrt(x^3y)+sqrt(xy^3)) \ (x>0,y>0)`

`P=[(sqrtx+sqrty)/sqrt(xy) . 2/(sqrtx+sqrty)+1/x+1/y] :(xsqrtx+ysqrtx+xsqrty+ysqrty)/(xsqrt(xy)+ysqrt(xy))`

`P=[2/sqrt(xy)+1/x+1/y]:(sqrtx(x+y)+sqrty(x+y))/(sqrt(xy)(x+y))`

`P=(2sqrt(xy)+y+x)/(xy):((x+y)(sqrtx+sqrty))/(sqrt(xy)(x+y))`

`P=(sqrtx+sqrty)^2/(xy) : (sqrtx+sqrty)/sqrt(xy)`

`P=(sqrtx+sqrty)^2/(xy) . sqrt(xy)/(sqrtx+sqrty)`

`P=(sqrtx+sqrty)/sqrt(xy)`

`b)`

Với `xy=16`, ta có `sqrt(xy)=4`

`P=(sqrtx+sqrty)/4`

Áp dụng bất đẳng thức Cauchy, ta có:

`sqrtx+sqrtyge2sqrt(sqrtx . sqrty) =2sqrt(sqrt(xy))`

`sqrtx+sqrtyge2sqrt4=4`

`Pge4/4=1`

Dấu "=" xảy ra khi và chỉ khi:

`{(sqrtx=sqrty),(xy=16):}=>{(x=y),(x^2=16):}=>x=y=4` (do `x,y>0`)

Vậy `P_(min)=1<=>x=y=4`

Bài `30:`

`a)`

`P=sqrt(x^3)/(sqrt(xy)-2y) -(2x)/(x+sqrtx-2sqrt(xy)-2sqrty) . (1-x)/(1-sqrtx) \ (xge0,y>0,xne4y,xne1)`

`P=(xsqrtx)/(sqrty(sqrtx-2sqrty)) - (2x)/(sqrtx(sqrtx+1)-2sqrty(sqrtx+1)) . ((1-sqrtx)(1+sqrtx))/(1-sqrtx)`

`P=(xsqrtx)/(sqrty(sqrtx-2sqrty))-(2x)/((sqrtx+1)(sqrtx-2sqrty)) . (1+sqrtx)`

`P=(xsqrtx)/(sqrty(sqrtx-2sqrty)) -(2x)/(sqrtx-2sqrty)`

`P=(xsqrtx-2xsqrty)/(sqrty(sqrtx-2sqrty))`

`P=(x(sqrtx-2sqrty))/(sqrty(sqrtx-2sqrty))`

`P=x/sqrty`

`b)`

Thay `y=625` vào `P` có:

`P=x/sqrt625=x/25`

Theo đề bài:

`P<0,2<=>x/25<1/5<=>x<25/5<=>x<5`

Kết hợp điều kiện `x` là số nguyên dương và `xne1`, ta có `x in {2,3,4}`

Vậy `x in {2,3,4}` thì `y=625,P<0,2`

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