Bài 8: Cộng, trừ đa thức một biến

13a+b+2c=0

=>b=-13a-2c

f(-2)=4a-2b+c=4a+c+26a+4c=30a+5c

f(3)=9a+3b+c=9a+c-39a-6c=-30a-5c

=>f(-2)*f(3)<=0

Lời giải:
a.

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

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

b.

$C(x)=4x-1=0$

$\Rightarrow x=\frac{1}{4}$

Vậy $x=\frac{1}{4}$ là nghiệm của $C(x)$

c.

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

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

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

a) f(x) = x³ - 2x² + x - x² - 5

= x³ + ( -2x² - x² ) + x - 5

= x³ - 3x² + x - 5

g(x) = -x³ + 4 + 3x² + 3x - 9

= -x³ + 3x² + 3x + ( 4 - 9)

= -x³ + 3x² + 3x - 5

b) f(x) + g (x)= x³ - 3x² + x - 5  - x³ + 3x² + 3x - 5

= ( x³ - x³ ) + ( -3x² + 3x² ) + ( x + 3x ) + ( -5 - 5 )

= 4x - 10 

f(x) - g(x) =  x³ - 3x² + x - 5 - ( -x³ + 3x² + 3x - 5)

=  x³ - 3x² + x - 5 + x³ - 3x² - 3x + 5

= ( x³ + x³ ) + ( -3x² - 3x² ) + ( x - 3x ) + ( -5 + 5)

= 2x³ - 6x² - 2x 

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

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

\(g\left(x\right)=-x^3+3x^2+3x+\left(4-9\right)\)

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

b,\(f\left(x\right)+g\left(x\right):\)

\(\Leftrightarrow f\left(x\right)+g\left(x\right)=x^3-2x^2+x-x^2-5+\left(-x^3+4+3x^2+3x-9\right)\)

\(\Leftrightarrow f\left(x\right)+g\left(x\right)=x^3-2x^2+x-x^2-5-x^3+4+3x^2+3x-9\)

\(\Leftrightarrow f\left(x\right)+g\left(x\right)=\left(x^3-x^3\right)+\left(-2x^2-x^2+3x^2\right)+\left(x+3x\right)+\left(-10\right)\)

 \(f\left(x\right)-g\left(x\right):\)

\(\Leftrightarrow f\left(x\right)-g\left(x\right)=x^3-2x^2+x-x^2-5-\left(-x^3+4+3x^2+3x-9\right)\)

\(\Leftrightarrow f\left(x\right)-g\left(x\right)=x^3-2x^2+x-x^2-5+x^3-4-3x^2-3x+9\)

\(\Leftrightarrow f\left(x\right)-g\left(x\right)=\left(x^3+x^3\right)+\left(-2x^2-x^2-3x^2\right)+\left(x-3x\right)+\left(9-4-5\right)\)

\(\Leftrightarrow f\left(x\right)-g\left(x\right)=2x^3-6x^2-2x\)

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

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

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

\(g\left(x\right)=-x^3+4+3x^2+3x-9\)

\(g\left(x\right)=-x^3+3x^2+3x+\left(4-9\right)\)

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

\(\text{b)f(x)+g(x)}=\left(x^3-3x^2+x-5\right)+\left(-x^3+3x^2+3x-5\right)\)

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

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

\(f\left(x\right)+g\left(x\right)=4x-10\)

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

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

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

\(f\left(x\right)-g\left(x\right)=2x^2-6x^2-2x\)

`M(x) = P(x) + Q(x)`

        `= ( 4x^3 - x^4 + x^2 - 5 + 2x ) + ( x^4 - 4x^3 + 5x^2 - 2x + 7 )`

       `= ( 4x^3 - 4x^3 ) + ( x^4 - x^4 ) + ( x^2 + 5x^2 ) + ( 2x-2x ) + ( 7-5 )`

       `= 6x^2 + 2`

Do `6x^2 \ge 0 AAx`

`=> 6x^2 + 2 \ge 2 \ne 0`

`=> M(x)` không có nghiệm `.`

`A(x) + Q(x) = P(x)` hay `A(x) = P(x) - Q(x)` 

                                              `=  ( 4x^3 - x^4 + x^2 - 5 + 2x ) - ( x^4 - 4x^3 + 5x^2 - 2x + 7 )`

                                             `=  ( 4x^3 - x^4 + x^2 - 5 + 2x ) -  x^4+ 4x^3 - 5x^2 + 2x - 7 `

                                             `= ( 4x^3 + 4x^3 ) + ( -x^4 - x^4 ) + ( x^2 - 5x^2 ) + ( 2x+2x)+(-5-7)` 

`= 8x^3 - 2x^4 - 4x^2 + 4x-12` 

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

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

\(M\left(x\right)=P\left(x\right)+Q\left(x\right)\)

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

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

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

\(M\left(x\right)=6x^2+2\)

\(\text{Đặt M(x)=0}\)

\(\Rightarrow6x^2+2=0\)

\(\Rightarrow6x^2\)       \(=0-2=-2\)

\(\Rightarrow x^2\)         \(=\left(-2\right):6=\dfrac{-1}{3}\)

\(\Rightarrow x\in\varnothing\)

\(\text{Vậy đa thức M(x) không có nghiệm}\)

\(A\left(x\right)+Q\left(x\right)=P\left(x\right)\)

\(\Rightarrow A\left(x\right)=P\left(x\right)-Q\left(x\right)\)

\(\Rightarrow A\left(x\right)=\left(-x^4+4x^3+x^2+2x-5\right)-\left(x^4-4x^3+5x^2-2x+7\right)\)

\(\Rightarrow A\left(x\right)=-x^4+4x^3+x^2+2x-5-x^4+4x^3-5x^2+2x-7\)

\(\Rightarrow A\left(x\right)=\left(-x^4-x^4\right)+\left(4x^3+4x^3\right)+\left(x^2-5x^2\right)+\left(2x+2x\right)+\left(-5-7\right)\)

\(\Rightarrow A\left(x\right)=-2x^4+8x^3-6x^2+4x-12\)

`|x|=3=>x=+-3`

`@` Thay `x=3` vào `A`. Ta có:

       `A=3.3^3-3+5=81-3+5=83`

`@` Thay `x=-3` vào `A`. Ta có:

     `A=3.(-3)^3-(-3)+5=-81+3+5=-73`

\(\left|x\right|=3\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

+) \(x=3\)

\(A=3.3^3-3+5=83\)

+) \(x=-3\)

\(A=3.\left(-3\right)^3-\left(-3\right)+5=-73\)

A=3.3³-3+5

A=3.27-3+5

A=81-3+5

A=78+5

A=83

\(\left|x\right|=3\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

+) \(x=3\)

\(A=3.3^3-3+5=83\)

+) \(x=-3\)

\(A=3.\left(-3\right)^3-\left(-3\right)+5=-73\)

\(\text{∆}=\left(-2\right)^2-4.2.3\)

\(=-20< 0\)

⇒ phương trình vô nghiệm