Đáp án: `a)A_{max}=\frac{13}{4}` khi `x=\frac{1}{2}`
`b)B_{max}=\frac{4}{3}` khi `x=\frac{4}{3};y=\frac{2}{3}`
Giải thích các bước giải:
a) Đặt $A=-x^2+x+3$
`=\frac{13}{4}-(x^2-2.x.\frac{1}{2}+\frac{1}{4})`
`=\frac{13}{4}-(x-\frac{1}{2})^2`
Do `(x-\frac{1}{2})^2≥0∀x`
`⇒-(x-\frac{1}{2})^2≤0`
`⇒A=\frac{13}{4}-(x-\frac{1}{2})^2≤\frac{13}{4}`
Dấu bằng xảy ra
`⇔(x-\frac{1}{2})^2=0`
`⇔x-\frac{1}{2}=0`
`⇔x=\frac{1}{2}`
b) Đặt $B=-x^2-y^2+xy+2x+2y$
$=-(x^2+y^2-xy-2x-2y)$
`=-[(x^2-xy+\frac{y^2}{4}-2x-y+1)+(\frac{3}{4}y^2-y+\frac{1}{3})-\frac{4}{3}]`
`=\frac{4}{3}-[(x-\frac{y}{2})^2-2(x-\frac{y}{2})+1]-\frac{3}{4}(y^2-\frac{4}{3}y+\frac{4}{9})`
`=\frac{4}{3}-(x-\frac{y}{2}-1)^2-\frac{3}{4}(y-\frac{2}{3})^2`
Do `(x-\frac{y}{2}-1)^2≥0∀x;y⇒-(x-\frac{y}{2}-1)^2≤0`
`(y-\frac{2}{3})^2≥0∀y⇒-\frac{3}{4}(y-\frac{2}{3})^2≤0`
`⇒-(x-\frac{y}{2}-1)^2-\frac{3}{4}(y-\frac{2}{3})^2≤0`
`⇒B=\frac{4}{3}-(x-\frac{y}{2}-1)^2-\frac{3}{4}(y-\frac{2}{3})^2≤\frac{4}{3}`
Dấu bằng xảy ra
`⇔(x-\frac{y}{2}-1)^2=(y-\frac{2}{3})^2=0`
Từ `(y-\frac{2}{3})^2=0⇔y-\frac{2}{3}=0⇔y=\frac{2}{3}`
`(x-\frac{y}{2}-1)^2=0⇔x-\frac{y}{2}-1=0`
`⇔x=\frac{1}{2}y+1=\frac{1}{2}.\frac{2}{3}+1=\frac{4}{3}`
