`1)A=x-6\sqrt{x}+1`
`A=x-6\sqrt{x}+9-8`
`A=(\sqrt{x}-3)^2-8`
Vì `(\sqrt{x}-3)^2>=0`
`=>A>=-8`
Dấu "=" xảy ra khi `\sqrt{x}-3=0<=>x=9`
`2)B=3x-8\sqrt{x}+2`
`B=3(x-8/3\sqrt{x})+2`
`B=3(x-2*\sqrt{x}*4/3+16/9)-16/3+2`
`B=3(\sqrt{x}-4/3)^2-10/3`
Vì `3(\sqrt{x}-4/3)^2>=0`
`=>B>=-10/3`
Dấu "=" xảy ra khi `\sqrt{x}-4/3=0<=>x=16/9`
`c)C=(\sqrt{x}-4)^2/\sqrt{x}(x>0)`
Vì `x>0=>\sqrt{x}>0`
Mà `(\sqrt{x}-4)^2>=0`
`=>C>=0`
Dấu "=" xảy ra khi `\sqrt{x}-4=0<=>x=16`
`d)D=(2x^2-3x-1)/(x-2)(x>2)`
`D=(2x^2+x-4x-2+1)/(x-2)`
`D=(x(2x+1)-2(2x+1)+1)/(x-2)`
`D=((x-2)(2x+1)+1)/(x-2)`
`D=2x+1+1/(x-2)`
`D=2x-4+1/(x-2)+5`
`D=2(x-2)+1/(x-2)+5`
Áp dụng bđt cosi ta có:
`2(x-2)+1/(x-2)>=2\sqrt{2}`
`=>D>=5+2\sqrt{2}`
Dấu "=" xảy ra khi `2/(x-2)=1/(x-2)`
`<=>(x-2)^2=1/2`
`<=>x-2=\sqrt{1/2}`
`<=>x=\sqrt{2}/2+2=(4+\sqrt{2})/2`
