`a)(x^2+x)^2-2(x^2+x)-15`
Đặt `x²+x=a`, khi đó ta được:
`a²-2a-15`
`=a²-5a+3a-15`
`=a(a-5)+3(a-5)`
`=(a-5)(a+3)`
`=(x²+x-5)(x²+x+3)`
`b)(x²+x+1)(x²+x+2)-12`
Đặt `x²+x+1=a` ,khi đó ta được:
`a(a+1)-12`
`=a²+a-12`
`=a²+4a-3a-12`
`=a(a+4)-3(a+4)`
`=(a+4)(a-3)`
`=(x²+x+1+4)(x²+x+1-3)`
`=(x²+x+5)(x²+x-2)`
`=(x²+x+5)(x²+2x-x-2)`
`=(x²+x+5)[x(x+2)-(x+2)]`
`=(x²+x+5)(x+2)(x-1)`
`c)x(x+4)(x+6)(x+10)+128`
`=[x(x+10)][(x+4)(x+6)]+128`
`=(x²+10x)(x²+6x+4x+24)+128`
`=(x²+10x)(x²+10x+24)+128`
Đặt `x²+10x=a`, khi đó ta được:
`a(a+24)+128`
`=a²+24a+128`
`=a²+8a+16a+128`
`=a(a+8)+16(a+8)`
`=(a+8)(a+16)`
`=(x²+10x+8)(x²+10x+16)`
`=(x²+10x+8)(x²+2x+8x+16)`
`=(x²+10x+8)[x(x+2)+8(x+2)]`
`=(x²+10x+8)(x+2)(x+8)`
`d)(2x^2-1)^4-5(2x^2-1)^2+4`
Đặt `2x^2-1=a` ,khi đó ta được:
`a^4-5a^2+4`
`=a^4-4a²-a²+4`
`=a²(a²-4)-(a²-4)`
`=(a²-4)(a²-1)`
`=(a²-2²)(a²-1²)`
`=(a+2)(a-2)(a+1)(a-1)`
`=(2x²-1+2)(2x²-1-2)(2x²-1+1)(2x²-1-1)`
`=(2x²+1)(2x²-3).2x²(2x²-2)`
`=4x²(2x²+1)(2x²-3)(x²-1)`
`=4x²(2x²+1)(2x²-3)(x+1)(x-1)`
