Cứu em với các ac mong a cj giúp

`A={3}/{x^2+2x+5}`

`={3}/{(x^2+2x+1)+4}`  

`={3}/{(x+1)^2+4}`  

Vì : `(x+1)^2>=0AAx`

`=>(x+1)^2+4>=4`

`=>{3}/{(x+1)^2+4}<=3/4` 

Hay `A<=3/4`

Dấu "=" xảy ra `<=>x=-1`

Vậy Max `A=3/4<=>x=-1`

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`B={14}/{9x^2+6x+4}`

`={14}/{[(3x)^2+2.3x.1+1^2]+3}`

`={14}/{(3x+1)^2+3}`

Vì : `(3x+1)^2>=0AAx`

`=>(3x+1)^2+3>=3`

`=>{14}/{(3x+1)^2+3}<= 14/3`

Hay `B<=14/3`

Dấu "=" xảy ra `<=>x=-1/3`

Vậy Max `B=14/3<=>x=-1/3`

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`C={13}/{4x^2+4x+3}`

`={13}/{[(2x)^2+2.2x.1+1^2]+2}`

`={13}/{(2x+1)^2+2}`

Vì : `(2x+1)^2>=0AAx`

`=>(2x+1)^2+2>=2`

`=>{13}/{(2x+1)^2+2} <= 13/2`

Hay `C<=13/2`

Dấu "=" xảy ra `<=>x=-1/2`

Vậy Max `C=13/2<=>x=-1/2`