Câu hỏi của Sách Giáo Khoa

Question

Cho các đa thức :

$f\left(x\right)=x^4-3x^2+x-1$

$g\left(x\right)=x^4-x^3+x^2+5$

Tìm đa thức $h\left(x\right)$ sao cho :

a) \(f\left(x\right)+h\left(x\right)=...

Lynk Lee · Accepted Answer

a) f (x)  + h (x) = g (x)

⇒h(x)=g(x)−f(x)⇒h(x)=g(x)−f(x)

h(x)=(x4−x3+x2+5)−(x4−3x2+x−1)h(x)=(x4−x3+x2+5)−(x4−3x2+x−1)

h(x)=x4−x3+x2+5−x4+3x2−x+1h(x)=−x3+4x2−x+6h(x)=x4−x3+x2+5−x4+3x2−x+1h(x)=−x3+4x2−x+6

b) f (x)  - h (x) = g (x)

⇒h(x)=f(x)−g(x)⇔h(x)=(x4−3x2+x−1)−(x4−x3+x2+5)⇒h(x)=f(x)−g(x)⇔h(x)=(x4−3x2+x−1)−(x4−x3+x2+5)

⇔h(x)=x4−3x2+x−1−x4+x3−x2−5⇔h(x)=x3−4x2+x−6Câu trả lời: a) f (x)  + h (x) = g (x)

⇒h(x)=g(x)−f(x)⇒h(x)=g(x)−f(x)

h(x)=(x4−x3+x2+5)−(x4−3x2+x−1)h(x)=(x4−x3+x2+5)−(x4−3x2+x−1)

h(x)=x4−x3+x2+5−x4+3x2−x+1h(x)=−x3+4x2−x+6h(x)=x4−x3+x2+5−x4+3x2−x+1h(x)=−x3+4x2−x+6

b) f (x)  - h (x) = g (x)

⇒h(x)=f(x)−g(x)⇔h(x)=(x4−3x2+x−1)−(x4−x3+x2+5)⇒h(x)=f(x)−g(x)⇔h(x)=(x4−3x2+x−1)−(x4−x3+x2+5)

⇔h(x)=x4−3x2+x−1−x4+x3−x2−5⇔h(x)=x3−4x2+x−6

いがつ · Answer

a. Ta có: f(x) + h(x) = g(x)

Suy ra: h(x) = g(x) – f(x) = (x4 – x3 + x2 + 5) – (x4 – 3x2 + x – 1)

= x4 – x3 + x2 + 5 – x4 + 3x2 – x + 1

= -x3 + 4x2 – x + 6

b. Ta có: f(x) – h(x) = g(x)

Suy ra: h(x) = f(x) – g(x) = (x4 – 3x2 + x – 1) – (x4 – x3 + x2 + 5)

= x4 – 3x2 + x – 1 – x4 + x3 – x2 – 5

= x3 – 4x2 + x – 6Câu trả lời: a. Ta có: f(x) + h(x) = g(x)

Suy ra: h(x) = g(x) – f(x) = (x4 – x3 + x2 + 5) – (x4 – 3x2 + x – 1)

= x4 – x3 + x2 + 5 – x4 + 3x2 – x + 1

= -x3 + 4x2 – x + 6

b. Ta có: f(x) – h(x) = g(x)

Suy ra: h(x) = f(x) – g(x) = (x4 – 3x2 + x – 1) – (x4 – x3 + x2 + 5)

= x4 – 3x2 + x – 1 – x4 + x3 – x2 – 5

= x3 – 4x2 + x – 6

Buithihavi · Answer

f(x) + h (x) = g (x)

(x^4 - 3x^2 + x - 1) + h(x) = x^4 - x^3 + x^2 + 5

h(x) = (x^4 - x^3 + x^2 +5) - (x^4 - 3x^2 + x - 1 )

h(x) = x^4 - x^3  + x^2 + 5 - x^4 + 3x^2 - x +1

h(x) = (x^4 - x^4 ) - x^3 + (x^2 + 3x^2) -x +1

h(x) = -x^3 + 4x^2 -x + 1Câu trả lời: f(x) + h (x) = g (x)

(x^4 - 3x^2 + x - 1) + h(x) = x^4 - x^3 + x^2 + 5

h(x) = (x^4 - x^3 + x^2 +5) - (x^4 - 3x^2 + x - 1 )

h(x) = x^4 - x^3  + x^2 + 5 - x^4 + 3x^2 - x +1

h(x) = (x^4 - x^4 ) - x^3 + (x^2 + 3x^2) -x +1

h(x) = -x^3 + 4x^2 -x + 1

Bài 8: Cộng, trừ đa thức một biến