Cái chỗ 1;1/2 là gì vậy bạn?

Bài 3: 

a) Đặt f(x)=0

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

b) Đặt f(x)=0

\(\Leftrightarrow x^2-7x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Bài 3:

c) Đặt f(x)=0

\(\Leftrightarrow x^2+2x+1=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

d) Đặt f(x)=0

\(\Leftrightarrow x^4+2=0\)

\(\Leftrightarrow x^4=-2\)(Vô lý)

a) N(x)= -2x³ + 5x² -12 +2x

M(x)= -x³ + 2,5x² - 0.5x -1

-

N(x)= -2x³ + 5x² + 2x - 12

=

A(x)=M(x) - N(x)= x³ - 2,5x² -2,5x +11

b) M(x) = -x³ + 2,5x² - 0,5x -1

+

N(x) = -2x³ + 5x² + 2x -12

=

B(x)= M(x) + N(x) = -3x³ + 7,5x² + 1,5x -13

⇒ Bậc của B(x) là 6

*Tính M(x) - N(x)

M(x)            = -x³ + 2,5x² - 0,5x - 1

N(x)            = -x³ + 2,5x² + 2x   - 6

------------------------------------

M(x) - N(x) =                    -2,5x + 5

=> M(x) - N(x) = A(x) = -2,5x + 5

Để đa thức A(x) có nghiệm => -2,5x + 5 = 0

=> -2,5x = -5

=> 2,5x = 5

=>  x = 2

Tính M(x) + N(x)  

M(x)          =   -x³ + 2,5x² - 0,5x - 1

N(x)          = -x³ + 2,5x² + 2x - 6 

---------------------------------------------

M(x) + N(x) = -2x³ + 5x² + 1,5x - 7

=> M(x) + N(x) = B(x) = -2x³ + 5x² + 1,5x - 7

Bậc của đa thức B(x) là 3

P/S : Cái dấu chấm đó là nhân hay phẩy?

a: \(M\left(x\right)=-2x^4-3x^2-7x-2\)

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

\(P\left(x\right)=M\left(x\right)+N\left(x\right)=-3x-7\)

Đặt P(x)=0

=>-3x-7=0

hay x=-7/3

b: Q(x)=N(x)-M(x)

\(=2x^4+3x^2+4x+5+2x^4+3x^2+7x+2\)

\(=4x^4+6x^2+11x+7\)

`a)P(x)=M(x)+N(x)`

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

         `=-3x-7`

Cho `P(x)=0`

`=>-3x-7=0`

`=>-3x=7`

`=>x=-7/3`

________________________________________________________

`b)Q(x)+M(x)=N(x)`

`=>Q(x)=N(x)-M(x)`

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

`=>Q(x)=4x^4+6x^2+11x-3`

`@` `\text {Ans}`

`\downarrow`

Ta có:

`A(x) = B(x)* Q(x) - x + 1`

`A(x) = x^3-2x^2+x`; `Q(x) = x - 1`

`<=> B(x) * (x - 1) - x + 1 = x^3 - 2x^2 + x`

`<=> B(x) * (x - 1) = x^3 - 2x^2 + x + x - 1`

`<=> B(x) * (x - 1) = x^3 - 2x^2 + 2x - 1`

`<=> B(x) = (x^3 - 2x^2 + 2x - 1) \div (x - 1)`

`<=> B(x) = x^2 - x + 1`

Vậy, `B(x) = x^2 - x + 1.`

A(x)=B(x)*Q(x)-x+1

=>x^3-2x^2+x=B(x)(x-1)-x+1

=>B(x)*(x-1)=x^3-2x^2+x+x-1=x^3-2x^2+2x-1

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

=>B(x)=x^2+x+1-2x

=>B(x)=x^2-x+1

Ta có: 

\(A\left(x\right)=B\left(x\right)\cdot Q\left(x\right)-x+1\)

\(\Leftrightarrow B\left(x\right)\cdot Q\left(x\right)=A\left(x\right)+x-1\)

\(\Leftrightarrow B\left(x\right)=\dfrac{A\left(x\right)+x-1}{Q\left(x\right)}\)

Mà: \(A\left(x\right)=x^3-2x^2+x\) và \(Q=x-1\) thay vào ta có:

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

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

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

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