1,
`x^2-10x-9y^2+25`
`=x^2-10x+25-9y^2`
`=x^2-2·x·5+5^2-9y^2`
`=(x-5)^2-(3y)^2`
`=(x-3y-5)(x+3y-5)`
2,
`A=3/(2x^2+2x+3)`
`=3/(2x^2+2x+1/2+5/2)`
`=3/((x√2+1/(√2))^2+5/2)`
Vì `(x√2+1/(√2))^2≥0∀x`
`→(x√2+1/(√2))^2+5/2≥5/2`
`→3/((x√2+1/(√2))^2+5/2)≤3/(5/2)=6/5`
Dấu `=` xảy ra `↔x√2+1/(√2)=0`
`↔x√2=-1/(√2)`
`↔x=-1/2`
Vậy `A_(max)=6/5↔x=-1/2`