Đáp án:
$\\$
`a,`
`A = 4x - x^2 + 3`
`↔ A = -x^2 + 4x + 3`
`↔ A = - x^2 + 4x -4 + 7`
`↔ A = - [x^2 - 4x + 4] + 7`
`↔ A = - [x^2 - 2 . 2x + 2^2] + 7`
`↔ A = - (x - 2)^2 + 7`
Với mọi `x` có : $(x-2)^2 \geqslant 0$
$↔ - (x-2)^2 \leqslant 0 ∀x$
$↔ - (x-2)^2 + 7 \leqslant 7 ∀ x$
$↔ A \leqslant 7 ∀ x$
`↔ max A = 7`
Dấu "`=`" xảy ra khi :
`↔x-2=0`
`↔x=0+2`
`↔x=2`
Vậy `max A=7 ↔ x=2`
$\\$
`b,`
`B = x^2 - 6x + 11`
`↔ B = x^2 - 6x + 9 + 2`
`↔ B = x^2 - 2 . 3x + 3^2 + 2`
`↔ B = (x-3)^2 +2`
Với mọi `x` có : $(x-3)^2 \geqslant 0$
$↔ (x-3)^2 + 2 \geqslant 2 ∀x$
$↔ B \geqslant 2 ∀ x$
`↔ min B=2`
Dấu "`=`" xảy ra khi :
`↔ x-3=0`
`↔x=0+3`
`↔x=3`
Vậy `min B=2 ↔ x=3`
$\\$
`c,`
`C = x^2 - 4x + y^2 - 8y + 6`
`↔ C = x^2 - 4x + y^2 - 8y + 4 + 2`
`↔ C = x^2 - 4x + 4 + y^2 - 8y + 16 - 14`
`↔ C = [x^2 - 4x + 4] + [y^2 - 8y + 16]-14`
`↔ C = [x^2-2 . 2x + 2^2] + [y^2 - 2 . 4y + 4^2] - 14`
`↔ C= (x-2)^2 + (y-4)^2 - 14`
Vợi mọi `x,y` có : \(\left\{ \begin{array}{l}(x-2)^2 \geqslant 0\\(y-4)^2 \geqslant 0\end{array} \right.\)
$↔ (x-2)^2 + (y-4)^2 \geqslant 0 ∀ x,y$
$↔ (x-2)^2 + (y-4)^2 - 14 \geqslant -14 ∀ x,y$
$↔ C \geqslant -14 ∀ x,y$
`↔ min C=-14`
Dấu "`=`" xảy ra khi :
`↔` \(\left\{ \begin{array}{l}x-2=0\\y-4=0\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=0+2\\y=0+4\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=2\\y=4\end{array} \right.\)
Vậy `min C=-14 ↔ x=2,y=4`
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`d,`
`D = x^2 - 8x + 19`
`↔ D = x^2 - 8x + 16 + 3`
`↔ D = x^2 - 2 . 4x + 4^2 +3`
`↔ D = (x-4)^2+3`
Với mọi `x` có : $(x-4)^2 \geqslant 0$
$↔ (x-4)^2 + 3 \geqslant 3 ∀ x$
$↔ D \geqslant 3 ∀ x$
`↔ min D=3`
Dấu "`=`" xảy ra khi :
`↔x-4=0`
`↔x=0+4`
`↔x=4`
Vậy `min D=3 ↔ x=4`
$\\$
`e,`
`E = -x^2 + 2x - 7`
`↔ E = -x^2 + 2x - 6 - 1`
`↔ E = - [x^2 - 2x + 1] - 6`
`↔ E = - [x^2 - 2 . 1x + 1^2] - 6`
`↔ E = - (x-1)^2-6`
Với mọi `x` có : $(x-1)^2 \geqslant 0$
$↔ - (x-1)^2 \leqslant 0 ∀x$
$↔ - (x-1)^2 -6 \leqslant -6 ∀x$
$↔ E \leqslant -6 ∀x$
`↔ max E = -6`
Dấu "`=`" xảy ra khi :
`↔x-1=0`
`↔x=0+1`
`↔x=1`
Vậy `max E=-6 ↔ x=1`
