tìm đa thức P và đa thức Q a, P+{3X2-4+5x}=x2-4x b, Q-14y4 +6×5-3 = -12y5+y4-1 19/08/2021 Bởi Amaya tìm đa thức P và đa thức Q a, P+{3X2-4+5x}=x2-4x b, Q-14y4 +6×5-3 = -12y5+y4-1
a) `P+ (3x^2 – 4+ 5x) = x^2 – 4x` `P= x^2 – 4x – (3x^2 -4 +5x)` `P= x^2 – 4x – 3x^2 + 4 – 5x` `P= (x^2 – 3x^2) – (4x+5x) +4` `P= -2x^2 – 9x +4` b) `Q- 14y^4 + 6x^5 – 3 = -12y^5 + y^4 -1` `Q= -12y^5 + y^4 -1 + 3 – 6x^5 + 14y^4` `Q= -12y^5 + (y^4 + 14y^4) – 6x^5 – (1-3)` `Q= -12y^5 + 15y^4- 6x^5 +2` Bình luận
Đáp án: $a/$ `P + (3x^2 – 4 + 5x) = x^2 – 4x` `-> P = x^2 – 4x – 3x^2 + 4 – 5x` `-> P = (x^2 – 3x^2) + (-4x – 5x) + 4` `-> P = -2x^2 – 9x + 4` $\\$ $b/$ `Q – 14y^4 + 6x^5 – 3 = -12y^5 + y^4 – 1` `-> Q = -12y^5 + y^4 – 1 + 14y^4 + 6x^5 – 3` `-> Q = -12y^5 + (y^4 + 14y^4) + (-1 – 3) + 6x^5` `-> Q = -12y^5 + 15y^4 – 4 + 6x^5` Bình luận
a) `P+ (3x^2 – 4+ 5x) = x^2 – 4x`
`P= x^2 – 4x – (3x^2 -4 +5x)`
`P= x^2 – 4x – 3x^2 + 4 – 5x`
`P= (x^2 – 3x^2) – (4x+5x) +4`
`P= -2x^2 – 9x +4`
b) `Q- 14y^4 + 6x^5 – 3 = -12y^5 + y^4 -1`
`Q= -12y^5 + y^4 -1 + 3 – 6x^5 + 14y^4`
`Q= -12y^5 + (y^4 + 14y^4) – 6x^5 – (1-3)`
`Q= -12y^5 + 15y^4- 6x^5 +2`
Đáp án:
$a/$ `P + (3x^2 – 4 + 5x) = x^2 – 4x`
`-> P = x^2 – 4x – 3x^2 + 4 – 5x`
`-> P = (x^2 – 3x^2) + (-4x – 5x) + 4`
`-> P = -2x^2 – 9x + 4`
$\\$
$b/$ `Q – 14y^4 + 6x^5 – 3 = -12y^5 + y^4 – 1`
`-> Q = -12y^5 + y^4 – 1 + 14y^4 + 6x^5 – 3`
`-> Q = -12y^5 + (y^4 + 14y^4) + (-1 – 3) + 6x^5`
`-> Q = -12y^5 + 15y^4 – 4 + 6x^5`