Phân tích đa tử thành nhân tử `x^4+4` `(x+1)(x+3)(x+3)(x+4)-1` `(x+1)(x+3)(x+3)(x+4)-24` free 27/11/2021 Bởi Adalynn Phân tích đa tử thành nhân tử `x^4+4` `(x+1)(x+3)(x+3)(x+4)-1` `(x+1)(x+3)(x+3)(x+4)-24` free
`a)x^4+4` `=x^4+4x^2+4-4x^2` `=(x^2+2)^2-(2x)^2` `=(x^2+2x+2)(x^2-2x+2)` `b)(x+1)(x+2)(x+3)(x+4)-1` `=[(x+1)(x+4)].[(x+2)(x+3)]-1` `=(x^2+5x+4)(x^2+5x+6)-1` Đặt `x^2+5x+4=a` `=a(a+2)-1` `=a^2+2a-1` `=(a-1)^2` `=(x^2+5x+4-1)^2` `=(x^2+5x-3)^2` `c)(x+1)(x+2)(x+3)(x+4)-24` `=[(x+1)(x+4)][(x+2)(x+3)]-24` `=(x^2+5x+4)(x^2+5x+6)-24` Đặt `x^2+5x+4=a` `=a(a+2)-24` `=a^2+2a-24` `=a^2+6a-4a-24` `=(a^2+6a)-(4a+24)` `=a(a+6)-4(a+6)` `=(a-4)(a+b)` `=(x^2+5x+10)(x^2+5x)` =.= Bình luận
`a)x^4+4`
`=x^4+4x^2+4-4x^2`
`=(x^2+2)^2-(2x)^2`
`=(x^2+2x+2)(x^2-2x+2)`
`b)(x+1)(x+2)(x+3)(x+4)-1`
`=[(x+1)(x+4)].[(x+2)(x+3)]-1`
`=(x^2+5x+4)(x^2+5x+6)-1`
Đặt `x^2+5x+4=a`
`=a(a+2)-1`
`=a^2+2a-1`
`=(a-1)^2`
`=(x^2+5x+4-1)^2`
`=(x^2+5x-3)^2`
`c)(x+1)(x+2)(x+3)(x+4)-24`
`=[(x+1)(x+4)][(x+2)(x+3)]-24`
`=(x^2+5x+4)(x^2+5x+6)-24`
Đặt `x^2+5x+4=a`
`=a(a+2)-24`
`=a^2+2a-24`
`=a^2+6a-4a-24`
`=(a^2+6a)-(4a+24)`
`=a(a+6)-4(a+6)`
`=(a-4)(a+b)`
`=(x^2+5x+10)(x^2+5x)`
=.=