Đáp án+Giải thích các bước giải:

`6) C = - ( |x + y - 1| + \sqrt{x/3 - 1} ) - 1/99` (ĐKXĐ: `x >= 3`)

Do `|x + y - 1| >= 0 ∀ y \in RR ; x >=3` và `\sqrt{x/3 - 1} >= 0 ∀ x >= 3`

`->  |x + y - 1| + \sqrt{x/3 - 1} >= 0`

`-> - ( |x + y - 1| + \sqrt{x/3 - 1}) <= 0`

`-> C <= 0 - 1/99 = -1/99`

Dấu "`=`" xảy ra khi `{(|x + y - 1| = 0),(\sqrt{x/3 - 1} = 0):}`

`-> {(x + y - 1 = 0),(x/3 - 1 = 0):}`

`-> {(y = -2),(x = 3):}`

Vậy `C_max = -1/99` khi `x = 3 ; y = -2`

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`7) C = -1/4 - (\sqrt{x - 3})/4 - |y + x - 6|` (ĐKXĐ: `x >= 3`)

Do `(\sqrt{x - 3})/4 >= 0 ∀ x > = 3`

`-> - (\sqrt{x - 3}/4 <= 0`

Lại có: `|y + x - 6| >= 0 ∀ y \in RR ; x >= 3`

`-> - |y + x - 6| <= 0`

`-> C <= -1/4 + 0 + 0 = -1/4`

Dấu "`=`" xảy ra khi: `{((\sqrt{x} - 3)/4 = 0),(|y + x - 6| = 0):}`

`-> {( \sqrt{x - 3} = 0),(y + x - 6 = 0):}`

`-> {( x = 3) , (y = 3):}`

Vậy `C_max = -1/4` khi `x = y = 3`

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`8) C = 3 + 2/(|x - 1| + (x + y - 1)^2 + 1)`

Do `|x - 1| >= 0 x \in RR ; (x + y - 1)^2 >= 0 x ; y \in RR`

`-> |x - 1| + (x + y - 1)^2 + 1 >= 0 + 0 + 1 = 2`

`-> 2/( |x - 1| + (x + y - 1)^2 + 1) <= 2/1 = 2`

`-> C <= 3 + 2 = 5`

Dấu "`=`" xảy ra khi: `{(|x - 1| = 0),((x + y - 1)^2 = 0):}`

`-> {(x - 1 = 0),(x + y - 1 = 0):}`

`-> {(x = 1),(y = 0):}`

Vậy `C_max = 5` khi `x = 1 ; y = 0`

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`9) C = - (x^2 - 9)^6 - 3/14 |y - 1/5| - 9`

Do `-(x^2 - 9)^6 <= 0 ∀ x \in RR` và `-3/14|y - 1/5| <= 0 ∀ y \in RR`

`-> C <= 0 + 0 - 9 = -9`

Dấu "`=`" xảy ra khi `{((x^2 - 9)^6 = 0),(|y - 1/5| = 0):}`

`-> {(x^2 - 9 = 0),(y - 1/5 = 0):}`

`-> {(x = ± 3),(y = 1/5):}`

Vậy `C_max = - 9` khi `(x ; y) = {(3 ; 1/5) ; (-3 ; 1/5)}`

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`10) C = 3 - 4/(|x + 2| + 3)`

Do `|x + 2| >= 0 ∀ x \in RR`

`-> |x + 2| + 3 >= 3`

`-> 4/(|x + 2| + 3) <= 4/3`

`-> -4/(|x + 2| + 3) >= -4/3`

`C >= 3 - 4/3 = 5/3`

Dấu"`=`" xảy ra khi `|x + 2| = 0`

`-> x + 2 = 0 `

`-> x = -2`

Vậy `C_min = 5/3` khi `x = -2`