`h)`

\(H=\left(x-2\right)\left(x-3\right)\left(x-6\right)\left(x+1\right)\\ =\left(x^2-5x+6\right)\left(x^2-5x-6\right)\\ =\left(x^2-5x\right)^2-36\)

`AA x` ta có `(x^2-5x)^2>=0`

`=>(x^2-5x)^2-36>=0-36`

hay `H>=-36`

Dấu `" = "` xảy ra khi `(x^2-5x)^2=0`

`x^2-5x=0`

`x(x-5)=0`

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy `Mi n H=-36` khi `x=0` hoặc `x=-5`

\(M=x^2+6y^2+2y+36-12x-2xy\\ =\left(x^2-2xy+y^2\right)-12x+12y+36+\left(5y^2-10y+5\right)-5\\ =\left[\left(x-y\right)^2-12\left(x-y\right)+36\right]+5\left(y^2-2y+1\right)-5\\ =\left(x-y-6\right)^2+5\left(y-1\right)^2-5\)

`AA x;y` ta có `{:((x-y-6)^2>=0),(5(y-1)^2>=0):}}`

`=>(x-y-6)^2+5(y-1)^2>=0`

`=>(x-y-6)^2+5(y-1)^2-5>=0-5`

hay `M>=-5`

Dấu `" = "` xảy ra khi \(\left\{{}\begin{matrix}\left(x-y-6\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-y-6=0\\y-1=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-7=0\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

Vậy `Mi n M=-5` khi `x=7;y=1`

`k)`

\(K=x^2+2y^2+2xy+2x+4y+2025\\ =\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)+2x+2y+2025\\ =[\left(x+y\right)^2+2\left(x+y\right)+1]+\left(y+1\right)^2+2024\\ =\left(x+y+1\right)^2+\left(y+1\right)^2+2024\)

`AA x;y` ta có : `{:((x+y+1)^2>=0),((y+1)^2>=0):}}`

`=>(x+y+1)^2+(y+1)^2>=0`

`=>(x+y+1)^2+(y+1)^2+2024>=0+2024`

hay `K>=2024`

Dấu `"="` xảy ra khi \(\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+y+1=0\\y+1=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0\\y=-1\end{matrix}\right.\)

Vậy `Mi n K =2024` khi `x=0; y=-1` 

\(A=\left(17^n+1\right)\left(17^n+2\right)\)

`AA n ` luôn có `17^n ; 17^n+1; 17^n+2` là `3` số liên tiếp

`=>17^n(17^n+1)(17^n+2) \vdots 3`

mà `17^n \cancel{vdots} 3` 

nên `(17^n+1)(17^n+2) \vdots 3`

hay `A \vdots 3(đpcm)`

`i)`

\(I=\left(x^2+4x+5\right)\left(x^2+4x+6\right)+3\\ =\left(x^2+4x+5\right)\left(x^2+4x+5+1\right)+3\\ =\left(x^2+4x+5\right)^2+\left(x^2+4x+5\right)+3\\ =\left(x^2+4x+5\right)^2+\left(x^2+4x+4\right)+1+3\\ =\left(x^2+4x+5\right)^2+\left(x+2\right)^2+4\)

Có `x^2+4x+5=x^2+4x+4+1`

hay `x^2+4x+5=(x+2)^2+1`

mà `(x+2)^2>=0 AA x => (x+2)^2+1>=1`

hay `x^2+4x+5>=1`

`AA x` ta có :`{:((x^2+4x+5)^2>=1),((x+2)^2>=0):}}`

`=>(x^2+4x+5)^2+(x+2)^2>=0+1`

`=>(x^2+4x+5)^2+(x+2)^2+4>=1+4`

hay `I>=5`

Dấu `"="` xảy ra khi \(\left\{{}\begin{matrix}\left(x^2+4x+5\right)^2=1\\\left(x+2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x^2+4x+5=1\left(x^2+4x+5>0\right)\\x+2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x^2+4x+4=0\\x=-2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=0\\x=-2\end{matrix}\right.\Rightarrow x=-2\)

Vậy `Mi n I=5` khi `x=-2`

bài `2:`

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\\ \left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=0\\ \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)=0\\ 2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)=-\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)

`AA a,b,c \ne 0` ta có :

`-(1/a^2 +1/b^2 +1/c^2)<=0`

hay `2(1/(ab)+1/(ac)+1/(bc))<=0`

`=>1/(ab)+1/(bc)+1/(ac)<=0(đpcm)`

Bài `4:`

`a)` Có `x=4/25` thỏa mãn ĐKXĐ `x>0,x\ne9`

nên Thay `x=4/25` vào `B` ta được:

\(B=\dfrac{\sqrt{\dfrac{4}{25}}-5}{\sqrt{\dfrac{4}{25}}}=\dfrac{\dfrac{2}{5}-5}{\dfrac{2}{5}}=-\dfrac{23}{2}\)

Vậy `B=-23/2` với `x=4/25`

`b)`\(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{x+9\sqrt{x}}{9-x}\left(ĐKXĐ:\text{x}>0,x\ne9\right)\\ =\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x-3}\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

Vậy `A=(\sqrt x)/(\sqrt x +3)` với `x>0, x\ne9`

`c)` Có `P=A*B`

hay \(P=\dfrac{\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}}\left(x>0,x\ne9\right)\\ =\dfrac{\sqrt{x}-5}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-8}{\sqrt{x}+3}\\ =1-\dfrac{8}{\sqrt{x}+3}\)

`AA x>0,x\ne9` ta có :

`\sqrt x>0`

`=>\sqrt x +3>0+3`

`=>8/(\sqrt x +3)<8/3`

`=>-8/(\sqrt x +3)> -8/3`

`=>1-8/(\sqrt x +3)>1-8/3`

hay `P> -5/3`

Dấu `=` xảy ra khi `\sqrt x=0`

                             `x=0(loại)`

Vậy không có giá trị nhỏ nhất của `P` thỏa mãn `x>0,x\ne 9`

\(\left(2x-5\right)^3=15^2-4\cdot5^2\\ \left(2x-5\right)^3=225-100\\ \left(2x-5\right)^3=125\\ \left(2x-5\right)^3=5^3\\ 2x-5=5\\ 2x=10\\ x=5\)