Bài 7:
`a)(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2`
Đặt `x^2+4x+8=a` ,khi đó ta được:
`a^2+3ax+2x^2`
`=a^2+2ax+ax+2x^2`
`=(a^2+ax)+(2ax+2x^2)`
`=a(a+x)+2x(a+x)`
`=(a+x)(a+2x)`
`=(x^2+4x+8+x)(x^2+4x+8+2x)`
`=(x^2+5x+8)(x^2+6x+8)`
`=(x^2+5x+8)(x^2+2x+4x+8)`
`=(x^2+5x+8)[(x^2+2x)+(4x+8)]`
`=(x^2+5x+8)[x(x+2)+4(x+2)]`
`=(x^2+5x+8)(x+2)(x+4)`
`b)(x^2+x+1)(x^2+x+2)-12`
Đặt `x^2+x+1=a` khi đó ta được:
`a(a+1)-12`
`=a^2+a-12`
`=a^2+4a-3a-12`
`=(a^2-3a)+(4a-12)`
`=a(a-3)+4(a-3)`
`=(a-3)(a+4)`
`=(x^2+x+1-3)(x^2+x+1+4)`
`=(x^2+x-2)(x^2+x+5)`
`=(x^2+2x-x-2)(x^2+x+5)`
`=[(x^2+2x)-(x+2)](x^2+x+5)`
`=[x(x+2)-(x+2)](x^2+x+5)`
`=(x+2)(x-1)(x^2+x+5)`
`c)(x^2+8x+7)(x^2+8x+15)+15`
Đặt `x^2+8x+11=a` ,khi đó ta được:
`(a-4)(a+4)+15`
`=a^2-4^2+15`
`=a^2-16+15`
`=a^2-1`
`=(a+1)(a-1)`
`=(x^2+8x+11+1)(x^2+8x+11-1)`
`=(x^2+8x+12)(x^2+8x+10)`
`=(x^2+2x+6x+12)(x^2+8x+10)`
`=[(x^2+2x)+(6x+12)](x^2+8x+10)`
`=[x(x+2)+6(x+2)](x^2+8x+10)`
`=(x+2)(x+6)(x^2+8x+10)`
`d)(x+2)(x+3)(x+4)(x+5)-24`
`=[(x+2)(x+5)][(x+3)(x+4)]-24`
`=(x^2+5x+2x+10)(x^2+4x+3x+12)-24`
`=(x^2+7x+10)(x^2+7x+12)-24`
Đặt `x^2+7x+11=a` ,khi đó ta được:
`(a-1)(a+1)-24`
`=a^2-1-24`
`=a^2-25`
`=a^2-5^2`
`=(a+5)(a-5)`
`=(x^2+7x+11+5)(x^2+7x+11-5)`
`=(x^2+7x+16)(x^2+7x+6)`
`=(x^2+7x+16)(x^2+6x+x+6)`
`=(x^2+7x+16)[(x^2+6x)+(x+6)]`
`=(x^2+7x+16)[x(x+6)+(x+6)]`
`=(x^2+7x+16)(x+6)(x+1)`
