Đáp án:
`A=5x-x^2`
`=-x^2+5x-25/4+25/4`
`=-(x^2-5x+25/4)+25/4`
`=-[x^2-2 . 5/2 .x +(5/2)^2]+25/4`
`=-(x-5/2)^2+25/4`
Vì `(x-5/2)^2>=0∀x`
`->-(x-5/2)^2<=0∀x`
`->-(x-5/2)^2+25/4<=25/4∀x`
`->A<=25/4∀x`
Dấu `'='` xảy ra `<=>x-5/2=0`
`<=>x=5/2`
Vậy `A_{max}=25/4` khi `x=5/2`
`B=x-x^2`
`=-x^2+x-1/4+1/4`
`=-(x^2-x+1/4)+1/4`
`=-[x^2 -2 . 1/2 .x +(1/2)^2]+1/4`
`=-(x-1/2)^2+1/4`
Vì `(x-1/2)^2≥0∀x`
`->-(x-1/2)^2≤0∀x`
`->-(x-1/2)^2+1/4≤1/4∀x`
`->B≤1/4∀x`
Dấu `'='` xảy ra `<=>x-1/2=0`
`<=>x=1/2`
Vậy `B_{max}=1/4` khi `x=1/2`
`C=4x-x^2+3`
`=-x^2+4x-4+7`
`=-(x^2-4x+4)+7`
`=-(x-2)^2+7`
Vì `(x-2)^2>=0∀x`
`->-(x-2)^2<=0∀x`
`->-(x-2)^2+7<=7∀x`
`->C<=7∀x`
Dấu `'='` xảy ra `<=>x-2=0`
`<=>x=2`
Vậy `C_{max}=7` khi `x=2`
`D=x^2+6x-11` ( câu này chỉ tìm min k có max)
Sửa `D=-x^2+6x-11`
`=-x^2+6x-9-2`
`=-(x^2+6x+9)-2`
`=-(x-3)^2-2`
Vì `(x-3)^2>=0∀x`
`->-(x-3)^2<=0∀x`
`->-(x-3)^2-2<=-2∀x`
`->D<=-2∀x`
Dấu `'='` xảy ra `<=>x-3=0`
`<=>x=3`
Vậy `D_{max}=-2` khi `x=3`
`E=5-8x-x^2`
`=-x^2-8x-16+21`
`=-(x^2+8x+16)+21`
`=-(x+4)^2+21`
Vì `(x+4)^2>=0∀x`
`->-(x+4)^2<=0∀x`
`->-(x+4)^2+21<=21∀x`
`->E≤21∀x`
Dấu `'='` xảy ra `<=>x+4=0`
`<=>x=-4`
Vậy `E_{max}=21` khi `x=-4`
______________________________
`A=x^2-6x+11`
`=x^2-6x+9+2`
`=(x^2-2.3.x+3^2)+2`
`=(x-3)^2+2`
Vì `(x-3)^2>=0∀x`
`->(x-3)^2+2>=2∀x`
`->A>=2∀x`
Dấu `'='` xảy ra `<=>x-3=0`
`<=>x=3`
Vậy `A_{min}=2`khi `x=3`
`B=x^2-20x+101`
`=x^2-20x+100+1`
`=(x^2-2.10.x+10^2)+1`
`=(x-10)^2+1`
Vì `(x-10)^2>=0∀x`
`->(x-10)^2+1>=1∀x`
`->B>=1∀x`
Dấu `'='` xảy ra `<=>x-10=0`
`<=>x=10`
Vậy `B_{min}=1` khi `x=10`
`C=x^2-2x+y^2+4y+8`
`=x^2-2x+1+y^2+4y+4+3`
`=(x^2-2x+1)+(y^2+2.2y+2^2)+3`
`=(x-1)^2+(y+2)^2+3`
Vì `(x-1)^2>=0∀x`
`(y+2)^2≥0∀y`
`->(x-1)^2+(y+2)^2≥0∀x;y`
`->(x-1)^2+(y+2)^2+3>=3∀x;y`
`->C>=3∀x;y`
Dấu `'='` xảy ra `<=>{(x-1=0),(y+2=0):}`
`<=>{(x=1),(y=-2):}`
Vậy `C_{min}=3` khi `(x;y)=(1;-2)`
