a, Ta có:

f(a + b) = 10(a + b)
f(a) + f(b) = 10a + 10b = 10(a+ b)

=> f(a + b) = f(a) + f(b)

b, f(x) = x² <=> 10x = x² 

                  <=> x = 10 hoặc x = 0
                

mk chiu mk ms hoc lop 6

Để \(f\left(x\right)=g\left(x\right)\Leftrightarrow2x+1=-x+3\Leftrightarrow x=\frac{2}{3}\)

Vậy \(x=\frac{2}{3}\)

để f(x)=g(x) thì suy ra 

2x+1=-x+3

suy ra x=\(\frac{2}{3}\)

a: \(F\left(3\right)=3\left(3-2\right)=3\cdot1=3\)

\(\left[F\left(\dfrac{2}{3}\right)\right]^2=\left[\dfrac{2}{3}\cdot\left(\dfrac{2}{3}-2\right)\right]^2\)

\(=\left[\dfrac{2}{3}\cdot\dfrac{-4}{3}\right]^2=\left(-\dfrac{8}{9}\right)^2=\dfrac{64}{81}\)

\(G\left(-\dfrac{1}{2}\right)=-\left(-\dfrac{1}{2}\right)+6=6+\dfrac{1}{2}=\dfrac{13}{2}\)

b: F(x)=0

=>x(x-2)=0

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

c: F(a)=G(a)

=>\(a\left(a-2\right)=-a+6\)

=>\(a^2-2a+a-6=0\)

=>\(a^2-a-6=0\)

=>(a-3)(a+2)=0

=>\(\left[{}\begin{matrix}a-3=0\\a+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-2\end{matrix}\right.\)

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

Bài 1:

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

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

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

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

      Giải:

Bài 1: lần lượt thay các giá trị của x, ta có:

_Y=f(-1)= -5.(-1)-1=4

_Y=f(0)= -5.0-1=1

_Y=f(1)= -5.1-1=-6

_Y=f(1/2)= -5.1/2-1=-7/2

Bài 2: 

a: f(-2)=7

f(-1)=5

f(0)=3