`1)`
`a)`
`(x^2+x)^2+4(x^2+x)=12`
`<=> (x^2+x)^2+4(x^2+x)-12=0`
`<=> (x^2+x)^2+4(x^2+x)+4-16=0`
`<=> (x^2+x+2)^2-4^2=0`
`<=> (x^2+x+2-4)(x^2+x+2+4)=0`
`<=> (x^2+x-2)(x^2+x+6)=0`
`<=> (x^2+2x-x-2)(x^2+x+6)=0`
`<=> [x(x+2)-(x+2)](x^2+x+6)=0`
`<=> (x-1)(x+2)(x^2+x+6)=0`
Mà `x^2+x+6=(x+1/2)^2+23/4>=23/4 ne 0`
`<=> (x-1)(x+2)=0`
`<=> [(x=1),(x=-2):}`
`b)`
`(x+5)(4-3x)-(3x+2)^2+(2x+1)^3=(2x-1)(4x^2+2x+1)`
`<=> 4x-3x^2+20-15x-(9x^2+12x+4)+8x^3+12x^2+6x+1=8x^3-1`
`<=> 20-11x-3x^2-9x^2-12x-4+8x^3+12x^2+6x+1-8x^3+1=0`
`<=> (8x^3-8x^3)+(12x^2-3x^2-9x^2)+(6x-12x-11x)+20-4+1+1=0`
`<=> -17x+18=0`
`<=> -17x=-18`
`<=> x=18/17`
`2)`
`-x^2+2x+5=-(x^2-2x-5)`
`=-(x^2-2x+1-6)=-[(x-1)^2-6]`
`=-(x-1)^2+6`
Vì : `(x-1)^2 ge 0 to -(x-1)^2 le 0 `
`to -(x-1)^2+6 le 6`
Dấu "=" xảy ra khi : `(x-1)^2=0`
`<=> x=1`
Vậy GTLN của biểu thức là : `6 <=> x=1`