Bài 7: Đa thức một biến

a) \(P\left(x\right)=5x^3-3x^2+8-3x^3+4x^5-3+8x+5x^2\)

\(P\left(x\right)=\left(5x^3-3x^3\right)-\left(3x^2-5x^2\right)+\left(8-3\right)+4x^5+8x\)

\(P\left(x\right)=2x^3+2x^2+5+4x^5+8x\)

b) Đa thức có bậc là: 5, hệ số cao nhất là 8, hệ số tự do là: 5

c) Sắp xếp theo lũy thừa giảm dần:

\(P\left(x\right)=2x^3+2x^2+5+4x^5+8x\)

\(P\left(x\right)=4x^5+2x^3+2x^2+8x+5\)

`(2x-3)(x^2-5+1)`

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

`= 2x*x^2+2x*(-5)+2x-[3*x^2+3*(-5)+3]`

`= 2x^3-10x+2x-(3x^2-15+3)`

`= 2x^3-10x+2x-3x^2+15-3`

`= 2x^3-3x^2-8x+12`

a. \(x\left(x-5\right)-\left(x-5\right)\left(3x+2\right)=0\)

\(\Rightarrow\left(x-5\right)\left[x-\left(3x+2\right)\right]=0\)

\(\Rightarrow\left(x-5\right)\left(x-3x-2\right)=0\)

\(\Rightarrow\left(x-5\right)\left(-2x-2\right)=0\)

\(\Rightarrow-2\left(x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-2\ne0\\x-5=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-2\ne0\\x=5\\x=-1\end{matrix}\right.\)

b. \(2\left(0,3x-0,5x\right)-0,5\left(3-x\right)=0\)

\(\Rightarrow-0,4x-1,5+0,5x=0\)

\(\Rightarrow0,1x-1,5=0\)

\(\Rightarrow0,1x=1,5\)

\(\Rightarrow x=\dfrac{1,5}{0,1}=15\)

c. \(\left(x-1\right)-\left(x-2\right)=\left(-x-40\right)-8\)

\(\Rightarrow x-1-x+2=-x-40-8\)

\(\Rightarrow1=-x-48\)

\(\Rightarrow1+48=-x\)

\(\Rightarrow x=-49\)

`@``dn10`

`a,`

`P(x)=-2x^5-3x^4+2x^5+2x-0,6`

`P(x)=(-2x^5+2x^5)-3x^4+2x-0,6`

`P(x)=-3x^4+2x-0,6`

`b,`

Thay `x=1` vào đa thức `B(x)`

`B(1)=-4*1^3+6*1-4=-4*1+6-4=-4+6-4=2-4=-2`

a: =-2x^5+2x^5+3x^4+2x-0,6

=3x^4+2x-0,6

b: B(1)=-4+6-4=-8+6=-2

=-2x^3+6x^2-x

`-2x.(x^2-3x+0,5)`

`=(-2x.x^2)+[-2x.(-3x)]+(-2x.0,5)`

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

1: (3x+2)(x+2)(2x-1)

=(3x^2+6x+2x+4)(2x-1)

=(3x^2+8x+4)(2x-1)

=6x^3-3x^2+16x^2-8x+8x-4

=6x^3+13x^2-4

2: (5x+1)(x-1)+3x(2x+2)

=5x^2-5x+x-1+6x^2+6x

=11x^2+10x-1

3: 4x(2x+1)(x-1)+(x+5)(x-3)

=4x(2x^2-2x+x-1)+x^2+2x-15

=8x^3-4x^2-4x+x^2+2x-15

=8x^3-3x^2-2x-15

4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)

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

=2x^3-8x-x^2+4+3x^2-4x+1

=2x^3+2x^2-12x+5

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

b: 

Sửa đề: 6x^4-4x^3+3x-2/3x-2

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

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

Ta có:

\(x-3=0\Rightarrow x=3\)

\(\Rightarrow\)Nghiệm của đa thức C(x) là 3

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

\(\Rightarrow x^3-3x^2-3x-1\) chia hết \(x^2+x+1\) khi \(3⋮x^2+x+1\)

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

Mà x nguyên dương \(\Rightarrow x^2+x+1\ge1^2+1+1=3\) (2)

(1);(2) \(\Rightarrow x^2+x+1=3\)

\(\Rightarrow x=1\)

Ta có:

\(P\left(x\right)=2x^3+5x^4+x^2-x^3-3x^4+2022+3x^2-x^3\)

\(P\left(x\right)=\left(5x^4-3x^4\right)+\left(2x^3-x^3-x^3\right)+\left(x^2+3x^2\right)+2022\)

\(P=2x^4+4x^2+2022\)