cho 2 đa thức:
P(x)= -3x^2+2x+1
Q(x)=-3x^2+x-2
a/ tính:
p(-2);p(-1);p(0);p(1);p(2)
q(-2);q(-1);q(1);q(2)q(0)
b/tính:
p(x)-q(x)
p(x)+q(x)
q(x)-p(x)
c/ tìm x để: p(x)=q(x)
cho 2 đa thức: P(x)= -3x^2+2x+1 Q(x)=-3x^2+x-2 a/ tính: p(-2);p(-1);p(0);p(1);p(2) q(-2);q(-1);q(1);q(2)q(0) b/tính: p(x)-q(x) p(x)+q(x) q(x)-p(x) c/
By Alice
Đáp án:
`a,`
`P (x) = -3x^2 + 2x + 1`
Có : `P (-2) = -3 . (-2)^2 + 2 . (-2) + 1`
`-> P (-2) = -3 . 4 – 4 + 1`
`-> P (-2) = -12 – 4 + 1 = -16 + 1`
`-> P (-2)= -15`
Có : `P (-1) = -3 . (-1)^2 + 2 . (-1) + 1`
`-> P (-1) = -3 . 1 – 2 + 1`
`-> P (-1)=-3 – 2 + 1 = -5 + 1`
`-> P (-1) = -4`
Có : `P (0) = -3 . 0^2 + 2 . 0 + 1`
`-> P (0) = -3 . 0 + 0 + 1`
`-> P (0) = 1`
Có : `P (1) = -3 . 1^2 + 2 .1 + 1`
`-> P (1) = -3 . 1 + 2 + 1`
`-> P (-1)= -3 + 2 + 1 = -1 + 1`
`-> P (-1) = 0`
Có : `P (2) = -3 . 2^2 + 2 . 2 + 1`
`-> P (2) =-3 . 4 + 4 + 1`
`-> P (2) = -12 + 4 + 1 = -8 + 1`
`->P (-2) = -7`
$\\$
`Q (x)=-3x^2 + x – 2`
Có : `Q (-2) =-3 . (-2)^2 + (-2) – 2`
`-> Q (-2) = -3 . 4 – 2 – 2 = -12 – 2 – 2`
`-> Q (-2) = -16`
Có : `Q (-1) = -3 . (-1)^2 + (-1) – 2`
`-> Q (-1) = -3 . 1 – 1 – 2 = -3 – 1 – 2`
`-> Q (-1) = -6`
Có : `Q (1) = -3 . 1^2 + 1 – 2`
`-> Q (1) = -3 . 1 + 1 – 2 = -3 + 1 – 2`
`-> Q (1) = -2 – 2`
`-> Q (1) = -4`
Có : `Q (2) = -3 . 2^2 + 2 – 2`
`-> Q (2) = -3 . 4 + 2 – 2 = -12 + 2 – 2`
`-> Q (2) = -10 – 2 = -12`
Có : `Q (0) = -3 . 0^2 + 0 – 2`
`-> Q (0) = -3 – 2`
`-> Q (0) = -5`
$\\$
$\\$
$b,$
`P (x) – Q (x) = -3x^2 + 2x + 1 + 3x^2 – x + 2`
`-> P (x) – Q (x) = (-3x^2 + 3x) + (2x – x) + (1 + 2)`
`-> P (x) – Q (x) = x + 3`
$\\$
`P (x) + Q (x) = -3x^2 + 2x + 1 – 3x^2 + x – 2`
`-> P (x) + Q (x) = (-3x^2 – 3x^2) + (2x + x) + (1 – 2)`
`-> P (x) + Q (x) = -6x^2 + 3x – 1`
$\\$
`Q (x) – P (x) = -3x^2 + x – 2 + 3x^2 – 2x – 1`
`-> Q (x) – P (x) = (-3x^2 + 3x^2) + (x – 2x) + (-2 – 1)`
`-> Q (x) – P (x) = -x – 3`
$\\$
$\\$
$c,$
`P (x) = Q (x)`
`-> -3x^2 + 2x + 1 = -3x^2 + x – 2`
`-> -3x^2 + 2x + 1 + 3x^2 – x + 2 = 0`
`-> x + 3 = 0`
`-> x = -3`
Vậy `x = -3` để `P (x) = Q (x)`
`\text{a)}`
Ta có :
`P(x) = -3x^2 + 2x +1`
$*$
`P(-2) = -3 . (-2)^2 + 2 . (-2) +1`
`P(-2) = -3 . 4 – 4 +1`
`P(-2) = -12 -4 +1`
`P(-2) = -15`
$*$
`P(-1) = -3 . (-1)^2 + 2 . (-1) + 1`
`P(-1) = -3 . 1 – 2 +1`
`P(-1) = -3 – 2 +1`
`P(-1) = -4`
$*$
`P(0) = -3 . 0^2 + 2 . 0 +1`
`P(0) = 0 + 0 +1`
`P(0) =1`
$*$
`P(1) = -3 . 1^2 + 2 . 1 +1`
`P(1) = -3 + 2 +1`
`P(1) = 0`
$*$
`P(2) = -3x^2 + 2x +1`
`P(2) = -3 . 2^2 + 2 . 2 + 1`
`P(2) = – 3 . 4 + 4 + 1`
`P(2) = -1`
`Q(x) = -3x^2 + x -2`
$*$
`Q(-2) = -3 . (-2)^2 + (-2) – 2`
`Q(-2) = -12 -4`
`Q(-2) = -16`
$*$
`Q(-1) = -3 . (-1)^2 + (-1) -2`
`Q(-1) = -3 – 3`
`Q(-1) = -6`
$*$
`Q(1) = -3 . 1^2 + 1 -2`
`Q(1) = -3 +1 -2`
`Q(1) =-4`
$*$
`Q(2) = -3 . 2^2 + 2 -2`
`Q(2) = -3 . 4 `
`Q(2) = -12`
$*$
`Q(0) = -3 . 0^2 + 0 -2`
`Q(0) = -2`
`\text{b)}`
Ta có :
$*$
`P(x) – Q(x) = (-3x^2+2x+1) – (-3x^2 +x-2)`
`P(x) – Q(x) = (-3x^2 +3x^2) + (2x – x) + (1+2)`
`P(x) – Q(x) = x +3`
$*$
`P(x)+ Q(x) = (-3x^2+2x +1) + (-3x^2+x-2)`
`P(x) + Q(x) = (-3x^2 – 3x^2) + (2x+x) + (1 -2)`
`P(x) + Q(x) = -6x^2 +3x -1 `
$*$
`Q(x) – P(x) = (-3x^2+x – 2) – (-3x^2+2x+1)`
`Q(x) – P(x) = (-3x^2+3x^2) + ( x -2 x) + (-2 -1)`
`Q(x) – P(x) = -3-x`
`\text{c)}`
Để `P(x) = Q(x)`
`=> -3x^2 + 2x +1 = -3x^2 + x -2`
`=> (-3x^2 + 2x +1) – (-3x^2 +x – 2) = 0`
`=> x +3 = 0` ( Kết quả tính ở phần `\text{b)}` )
`=> x = -3`
Vậy `x = -3` thì `P(x) = Q(x)`