Bài 5
Giải: Ta có: [tex]P=x^{2}+4x-2\left | x+2 \right |+2025[/tex]
[tex]P=(x^{2}+4x+4)-2\left | x+2 \right |+2021[/tex]
[tex]=(x+2)^{2}-2\left | x+2 \right |+2021[/tex]
[tex]= \left [ \left | x+2 \right | \right ]^{2}-2\left [ x+2 \right ]+1+2020[/tex]
[tex]=\left [ \left | X+2 \right |-1 \right ]^{2}+2020\geq 2020[/tex]
Vì [tex]\left [ \left | X+2 \right |-1 \right ]^{2}\geq 0[/tex]
Dấu "=" xảy ra khi [tex]\left | x+2 \right |-1=0[/tex]
[tex]\Leftrightarrow \left | x+2 \right |=1[/tex]
[tex]\Leftrightarrow x+2=-1[/tex] hoặc [tex]x+2=1[/tex]
[tex]\Leftrightarrow x=-3[/tex] hoặc [tex]x=-1[/tex]
Vậy GTLN là 2020 khi [tex]x=-3[/tex] hoặc [tex]x=-1[/tex]