\(b,f\left(1\right)=\dfrac{-2}{5}.1=-\dfrac{2}{5}\\f\left(-2\right) =\dfrac{-2}{5}\left(-2\right)=\dfrac{4}{5}\)

c: Ở hai hàm số trên, nếu lấy biến x cùng một giá trị thì f(x) sẽ nhỏ hơn g(x) 3 đơn vị

a: f(-1)=4

f(2)=1

\(a,f\left(-1\right)=3-\left(-1\right)=4\\ f\left(2\right)=3-2=1\\ b,y=5\Rightarrow3-x=5\Rightarrow x=-2\\ y=2\Rightarrow3-x=2\Rightarrow x=1\\ y=-1\Rightarrow3-x=-1\Rightarrow x=4\)

a: f(-2)=3-(-2)=3+2=5

f(-1)=3-(-1)=4

b: y=5 thì 3-x=5

hay x=-2

a,f[-2]=3-[-2]=5

   f[-1]=3-[-1]=4

b,y=f[x]=3-x

  =>5=3-x

  =>x=3-5

  =>x=-2

Ta có: \(y=f\left(x\right)=3-x\)

a) \(f\left(-2\right)\Rightarrow3-\left(-2\right)=5\)

\(f\left(-1\right)\Rightarrow3-\left(-1\right)=4\)

b) \(y=5\Rightarrow3-x=5\Rightarrow x=-2\)

Câu 1: 

a) 

b) Ta có: f(x)=8

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

Vậy: Để f(x)=8 thì \(x\in\left\{2;-2\right\}\)

Ta có: \(f\left(x\right)=6-4\sqrt{2}\)

\(\Leftrightarrow2x^2=6-4\sqrt{2}\)

\(\Leftrightarrow x^2=3-2\sqrt{2}\)

\(\Leftrightarrow x=\sqrt{3-2\sqrt{2}}\)

hay \(x=\sqrt{2}-1\)

Vậy: Để \(f\left(x\right)=6-4\sqrt{2}\) thì \(x=\sqrt{2}-1\)

giúp em ạ =))))

a: f(1)=2

f(-1)=-2

f(1/2)=1

f(-1/2)=-1

\(f\left(0\right)=3.0-1=-1\\ f\left(\dfrac{1}{3}\right)=3.\dfrac{1}{3}-1=0\)

f(0)=-1

f(1/3)=0