`b)ĐKXĐ: x>=0`
`A=2x-3\sqrt{x}-7`
`A=2(x-3/2\sqrt{x}-7/2)`
`A=2[(\sqrt{x})^2-2.\sqrt{x}.(3)/4+(3/4)^2-65/16]`
`A=2(\sqrt{x}-3/4)-65/8>=-65/8`
Dấu `=` xảy ra
`<=>\sqrt{x}-3/4=0`
`<=>\sqrt{x}=3/4`
`<=>x=9/16(TM)`
Vậy `minA=-65/8` khi `x=9/16`
`c)ĐKXĐ: x>0`
Áp dụng bất đẳng thức `Cosi` ta có:
`4/\sqrt{x}+\sqrt{x}>=2\sqrt{4/(\sqrt{x}).\sqrt{x}}=4`
Dấu `=` xảy ra
`<=>4/\sqrt{x}=\sqrt{x}`
`<=>x=4(TM)`
`=>A>=4+1=5`
Vậy `minA=5` khi `x=4`