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Bài `4:`

`a)`

Ta có`:`

`|2x+1|>=0`

`3|2x+1|>=0`

`3|2x+1|>=-4`

Dấu "`=`" xảy ra khi`:`

`2x+1=0`

`2x=-1`

`x=(-1)/2`

Vậy `minA=-4` khi `x=(-1)/2`

`b)`

Ta có`:`

`{((x-2)^2022 >=0),((y+1)^2024 >=0):}`

Do đó`:`

`(x-2)^2022 +(y+1)^2024 >=0`

`(x-2)^2022 + (y+1)^2024 -2023 >= -2023`

Dấu "`=`" xảy ra khi`:`

`{((x-2)^2022 =0),((y+1)^2024 =0):}`

`{(x-2=0),(y+1=0):}`

Suy ra `{(x=2),(y=-1):}`

Vậy `minB=-2023` khi `(x,y)=(2,-1)`

`c)`

Ta có`:`

`D=|2x-1|+|5-2x|`

Do`: {(|2x-1| >= 2x-1),(|5-2x|>=5-2x):}`

Nên`:`

`D>=2x-1+5-2x`

`D>= 4`

Dấu "`=`" xảy ra khi`:`

`{(2x-1 >=0),(5-2x >=0):}`

`{(2x>=1),(2x<=5):}`

`{(x >=1/2),(x<=5/2):}`

Vậy `minD=4` khi `1/2 <= x <=5/2`

Bài `5:`

`a)`

Ta có`:`

`(2x-3)^4 >=0`

`(2x-3)^4 +5 >=5`

`3/((2x-3)^4 +5) <= 3/5`

Dấu "`=`" xảy ra khi`:`

`(2x-3)^4=0`

`2x-3=0`

`2x=3`

`x=3/2`

Vậy `maxA=3/5` khi `x=3/2`

`b)`

Ta có`:`

`-1/2(x-1)^2 <=0`

Lại có`:`

`-2|y+1| <= 0`

Do đó`:`

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

`-1/2 (x-1)^2 -2|y+1| -11 <= -11`

Dấu "`=`" xảy ra khi`:`

`{((x-1)^2=0),(y+1=0):}`

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

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

Vậy `maxB=-11` khi `(x,y)=(1,-1)`