phân tích đa thức thành nhân tử (x+1)(x+2)(x+3)(x+4)+1 10/08/2021 Bởi Kaylee phân tích đa thức thành nhân tử (x+1)(x+2)(x+3)(x+4)+1
`(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` $\text{Đặt: $x^2$ + 5x + 4 = y}$ `⇒y(y+2)+1` `=y^2+2y+1` `=(y+1)^2` $\text{Quay trở lại ta có:}$ `⇒(x^2+5x+4+1)^2` `=(x^2+5x+5)^2` Bình luận
`(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``= ( x^2 + 5x + 5 -1)(x^2 + 5x + 5 + 1 ) + 1``= ( x^2 + 5x + 5)^2 – 1 + 1``= ( x^2 + 5x + 5)^2` Bình luận
`(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`
$\text{Đặt: $x^2$ + 5x + 4 = y}$
`⇒y(y+2)+1`
`=y^2+2y+1`
`=(y+1)^2`
$\text{Quay trở lại ta có:}$
`⇒(x^2+5x+4+1)^2`
`=(x^2+5x+5)^2`
`(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`
`= ( x^2 + 5x + 5 -1)(x^2 + 5x + 5 + 1 ) + 1`
`= ( x^2 + 5x + 5)^2 – 1 + 1`
`= ( x^2 + 5x + 5)^2`