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

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

a: f(-1/3)=-1/27-2/9-1+1=-1/27-6/27=-7/27

g(-2)=-8+4+10+3=17-8=9

b: f(x)-g(x)=x^3-2x^2+3x+1-x^3-x^2+5x-3

=x^2+8x-2

f(x)+g(x)

=x^3-2x^2+3x+1+x^3+x^2-5x+3

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

a: f(x) chia hết cho x^2+x+1

=>\(x^3+x^2+x+\left(a-1\right)x^2+\left(a-1\right)x+a-1-ax+b+1⋮x^2+x+1\)

=>-a=0 và b+1=0

=>a=0 và b=-1

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

\(=x+a+\dfrac{x+b+a}{x^2-1}\)

Để f(x) chia x^2-1 dư x+3 thì x+b+a=x+3

=>b+a=3

2: ĐKXĐ: x<>1

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

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

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

f'(x)=0

=>x^2-2x=0

=>x(x-2)=0

=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

1:

\(f\left(x\right)=\dfrac{1}{3}x^3-2\sqrt{2}\cdot x^2+8x-1\)

=>\(f'\left(x\right)=\dfrac{1}{3}\cdot3x^2-2\sqrt{2}\cdot2x+8=x^2-4\sqrt{2}\cdot x+8=\left(x-2\sqrt{2}\right)^2\)

f'(x)=0

=>\(\left(x-2\sqrt{2}\right)^2=0\)

=>\(x-2\sqrt{2}=0\)

=>\(x=2\sqrt{2}\)

`1,`

`f(x)+g(x)=(5+4-2+7)+(4x^4-2x^3+3x^2+4x-1)`

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

`=(5x^4+4x^4)-2x^3+(4x^2+4x^2)+(-2x+4x)+(7-1)`

`= 9x^4-2x^3+8x^2+2x+6`

Đề phải là `f(x)-g(x)` chứ nhỉ :v?

`f(x)-g(x)=(5+4-2+7)-(4x^4-2x^3+3x^2+4x-1)`

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

`= (5x^4-4x^4)+2x^3+(-2x-4x)+(4x^2-3x^2)+(7+1)`

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

`2,`

`a, (x+3)(x-1)`

`= x(x-1)+3(x-1)`

`= x*x+x*(-1)+3*x+3*(-1)`

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

`= x^2+2x-3`

`b, (4x+3)(x-2)`

`= 4x(x-2)+3(x-2)`

`= 4x*x+4x*(-2)+3*x+3*(-2)`

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

`c, (2x+3)(x+1)`

`= 2x(x+1)+3(x+1)`

`= 2x*x+2x*1+3*x+3*1`

`= 2x^2+2x+3x+3`

`= 2x^2+5x+3`

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

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

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

`= 5x^3-15x^2+5x-2x^2+6x-2`

`= 5x^3-17x^2+11x-2`