Bài 1:

Để \(F\left(x\right)=G\left(x\right)\) thì \(3x^2-8x+4=3x+4\)

\(\Leftrightarrow3x^2-11x=0\)

\(\Leftrightarrow x\left(3x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{11}{3}\end{matrix}\right.\)

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

câu 4: b, đề bài là tính giá trị của A tại x =-1/2;y=-1

Tk

Bài 2

a) F(x)-G(x)+H(x)= \(x^3-2x^2+3x+1-\left(x^3+x-1\right)+\left(2x^2-1\right)\)

= \(x^3-2x^2+3x+1-x^3-x+1+2x^2-1\)

=  \(x^3-x^3-2x^2+2x^2+3x-x+1+1-1\)

=  2x + 1

b) 2x + 1 = 0

 2x = -1

 x=\(\dfrac{-1}{2}\)

Tk

Bài 3

a)

f(x) + g(x)

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

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

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

\(-x-2+2x^2\)

f(x) - g(x)

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

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

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

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

b)

Thay x = -1, ta có:

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

x = -2, ta có

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

\(2\cdot\left(-8\right)+6+4-8\) = -14

Ta có: \(f'\left(x\right)=\dfrac{3}{2\sqrt{3x+1}}\)

Do đó, \(f\left(1\right)=2,f'\left(1\right)=\dfrac{3}{4}\)

Vậy \(g\left(2\right)=f\left(1\right)+4\left(2^2-1\right)f'\left(1\right)=2+12\cdot\dfrac{3}{4}=11\)

a) f(-2)=2.(-2)+1=-4+1=-3

    f(a)=2a+1

b) g(-2)=3.(-2)-2= - 6-2=-8

   g(a)=3a-2

k đúng cho mk nha

a) f(-2) = -1; f(-1) = 0; f(0) = 1; f(2) = 3

g(-1) = 0,5; g(-2) = 2; g(0) = 0

b) f(x) = 2 ⇒ x = 1

g(x) = 2 ⇒ x = 2 hoặc x = -2

a,  

f(-1)=3*(-1)+1=-2

g(-1)=1-3*(-1)=4

b,

để f(x)=g(x)

thì 3*x+1=1-3*x

=> 3*x+3*x=1-1

     6*x=0

        x=0/6

         x=0

a) Thay f(x)=f(-1)=3.(-1)+1=-2

Thay g(x)=g(-1)=1-3.(-1)=4

=> f(x)+g(x)=-2+4=2

b) f(x)=g(x)=3x+1=1-3x

=> (3x+1)-(1-3x)=0

<=> 3x+1-1+3x=0

<=> 6x=0 => x=0