Bài 8: Cộng, trừ đa thức một biến

Em chưa làm được cả 2 bài hả em?

2:

a: C(x)=5x^6-9x^6+4x^4-6x^2-3x^2-6

=-4x^6+4x^4-9x^2-6

\(D\left(x\right)=-6x^5+3x^3+x^3-5x^2-4x^2+2x-7x+2\)

=-6x^5+4x^3-9x^2-5x+2

b: C(x): Bậc 0: 6; bậc 2: -9; bậc 4: 4; bậc 6: -4

D(x): bậc 5: -6; bậc 3: 4; bậc 2: -9; bậc 1: -5; bậc 0: 2

c: P(x)=C(x)-D(x)

=-4x^6+4x^4-9x^2-6+6x^5-4x^3+9x^2+5x-2

=-4x^6+6x^5+4x^4-4x^3+5x-8

Ta có: 

`M(x)+N(x)=2x^2-3x`

`(-2x^2+8x+4)+N(x)=2x^2-3x`

`-> N(x)=(2x^2-3x)-(-3x^2+8x+4)`

`N(x)=2x^2-3x+3x^2-8x-4`

`N(x)=(2x^2+3x^2)+(-3x-8x)-4`

`N(x)=5x^2-11x-4`

N(x)=2x^2-3x+2x^2-8x+4=4x^2-11x+4

1. \(\left(2x^2-3x+1\right)+\left(3x^2+2x-1\right)\)

\(=2x^2-3x+1+3x^2+2x-1\)

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

\(=5x^2-x\)

2. \(\left(4x^3-2x^2+3x\right)-\left(2x^3+3x^2-4x\right)\)

\(=4x^3-2x^2+3x-2x^3-3x^2+4x\)

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

\(=2x^3-5x^2+7x\)

3. \(\left(x^2-5x+6\right)+\left(-3x^2+2x-1\right)\)

\(=x^2-5x+6-3x^2+2x-1\)

\(=\left(x^2-3x^2\right)-\left(5x-2x\right)+\left(6-1\right)\)

\(=-2x^2-3x+5\)

4. \(\left(2x^3+5x^2-3x+1\right)-\left(x^3-2x^2+x-1\right)\)

\(=2x^3+5x^2-3x+1-x^3+2x^2-x+1\)

\(=\left(2x^3-x^3\right)+\left(5x^2+2x^2\right)-\left(3x+x\right)+\left(1+1\right)\)

\(=x^3+7x^2-4x+2\)

5. \(\left(3x^2+2x-4\right)+\left(4x^2-x+5\right)\)

\(=3x^2+2x-4+4x^2-x+5\)

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

\(=7x^2+x+1\)

6. \(\left(x^3-2x^2+5x-1\right)-\left(2x^3+3x^2-4x+2\right)\)

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

\(=\left(x^3-2x^3\right)-\left(2x^2+3x^2\right)+\left(5x+4x\right)-\left(1+2\right)\)

\(=-x^3-5x^2+9x-3\)

\(=3x^2\left(x^2+3x+1\right)\)

=3x^4+9x^3+3x^2

a: =2x^4-8x^2+10x+x^3-4x+5

=2x^4+x^3-8x^2+6x+5

b: =\(\left(-x^3-\dfrac{1}{2}x\right)\left(2x^3+x^2-1\right)\)

\(=-2x^6-x^5+x^3-x^4-x^3+\dfrac{1}{2}x\)

\(=-2x^6-x^5-x^4+\dfrac{1}{2}x\)

c: \(=-5x^4-4x^3+5x^2+25x^3+20x^2-25x\)

\(=-5x^4+21x^3+25x^2-25x\)

d: \(=\left(3x^2+9x+3\right)\left(2x-x^2\right)\)

\(=6x^3-3x^4+18x^2-9x^3+6x-3x^2\)

\(=-3x^4-3x^3+15x^2+6x\)

a: M(x)=3x^3-7x^2+4/5x-1/5

N(x)=3x^3-7x^2-x+3/2

b: M(x)+N(x)=6x^3-14x^2-1/5x+13/10

c: H(x)=M(x)-N(x)

=3x^3-7x^2+4/5x-1/5-3x^3+7x^2+x-3/2

=9/5x-17/10

`a, P(x)=15-4x^2+3x-2x^3`

`P(x)=-2x^3-4x^2+3x+15`

`Q(x)=5+4x^3+6x^2-5x`

`Q(x)=4x^3+6x^2-5x+5`

`b, P(x)+Q(x)= (-2x^3-4x^2+3x+15)+(4x^3+6x^2-5x+5)`

`P(x)+Q(x)=  -2x^3-4x^2+3x+15+4x^3+6x^2-5x+5`

`= (-2x^3+4x^3)-(4x^2-6x^2)+(3x-5x)+(15+5)`

`= 2x^3+2x^2-2x+20`

`c, P(x)-Q(x)= (-2x^3-4x^2+3x+15)-(4x^3+6x^2-5x+5)`

`P(x)-Q(x)= -2x^3-4x^2+3x+15-4x^3-6x^2+5x-5`

`= (-2x^3-4x^3)-(4x^2+6x^2)+(3x+5x)+(15-5)`

`= -6x^3-10x^2+8x+10`

a: A=x^3-2x^2+5x-1

B=x^3-3x^2+3x-2

P=A+B=2x^3-5x^2+8x-3

Q=A-B=x^2+2x+1

b: Bậc của P lớn hơn Q

c: Q(-1)=(-1)^2+2*(-1)+1=0

=>x=-1 là nghiệm của Q

\(1)A\left(x\right)+B\left(x\right)\)

\(=2x^2-3x-5+2x^2-5x-5\)

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

\(=4x^2-8x-10\)

\(2)B\left(x\right)-A\left(x\right)\)

\(=\left(2x^2-5x-5\right)-\left(2x^2-3x-5\right)\)

\(=2x^2-5x-5-2x^2+3x+5\)

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

\(=-2x\)

\(3)A\left(-1\right)\)

Thay \(x=-1\) vào \(A\left(x\right)=2x^2-3x-5\)

\(\Rightarrow A\left(-1\right)=2.\left(-1\right)^2-3.\left(-1\right)-5=2+3-5=0\)

=x^4+5x^3-5x^3+2x^2+10x-6x+1

=x^4+2x^2+4x+1