a) Ta có: \(y=f\left(x\right)=4x^2-5\)

\(\Rightarrow\left\{{}\begin{matrix}f\left(3\right)=4.3^2-5=31\\f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2-5=-4\end{matrix}\right.\)

b) Ta có: \(f\left(x\right)=-1\)

\(\Rightarrow4x^2-5=-1\)

\(\Leftrightarrow4x^2=4\)

\(\Leftrightarrow x^2=1\)

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

Vậy \(x\in\left\{1;-1\right\}\) thì \(f\left(x\right)=-1\)

c) \(\forall x\in R,f\left(x\right)=f\left(-x\right)\Leftrightarrow f\left(-x\right)=4.\left(-x\right)^2-5=4x^2-5=f\left(x\right)\)

Vậy \(\forall x\in R\) thì \(f\left(x\right)=f\left(-x\right)\)

\(a.f\left(3\right)=4.3^2-5=31.\\ f\left(\dfrac{-1}{2}\right)=4.\left(\dfrac{-1}{2}\right)^2-5=-4.\)

\(b.f\left(x\right)=-1.\Rightarrow4x^2-5=-1.\\ \Leftrightarrow4x^2=4.\Leftrightarrow x^2=1.\\ \Leftrightarrow x=\pm1.\)

\(c.f\left(x\right)=f\left(-x\right).\\ \Rightarrow4x^2-5=4\left(-x\right)^2-5.\\ \Leftrightarrow4x^2-5=4x^2-5.\)

\(\Leftrightarrow0x=0\) (luôn đúng).

Vậy với mọi x ∈ R thì f (x)= f (-x).

a) Do \( y=f(x)=4x² - 5 \) nên : 

\(+) f(3) = 4 . 3^2 - 5 = 4 . 9 - 5 = 36 - 5 = 31 \) 

\(+) f(\dfrac{1}{2}) = 4 . (\dfrac{1}{2})^2 - 5 = 4 . \dfrac{1}{4} - 5 = 1 - 5 = -4 \) 

Vậy : \(f(3) = 31 ; f(\dfrac{1}{2}) = -4 \) 

b) Do \(f(x) = -1 \) 

Mà \(f(x) = 4x^2 - 5 \) 

\(=> \) \(4x^2 - 5 = -1 \)

\(=> 4x^2 = -1 + 5 \) 

\(=> 4x^2 = 4 \) 

\(=> x^2 = 1 \) \(= 1^2 = ( -1)^2 \) 

\(=> x \) ∈ { -1 ; 1 }

Vậy với \(f(x) = -1 \) thì x ∈ { -1 ; 1 } 

c) Ta có : Do \(x^2 = ( -x )^2 \) 

\(=> \) \(4x^2 = 4(-x)^2 \) 

\(=> 4x^2 - 5 = 4( -x )^2 - 5 \) 

\(=> f(x) = f(-x) \)

Vậy với mọi x ∈ R thì \(f(x) = f(-x)\) 

\(y=f\left(x\right)=4x^2-9\)

a, \(f\left(-2\right)=4.\left(-2\right)^2-9\)

\(=16-9\)

\(=7\)

 \(f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2-9\)

                 \(=4.\dfrac{1}{4}-9\)

\(=1-9\)

\(=-8\)

b, \(f\left(x\right)=-1\Rightarrow4x^2-9=-1\)

\(\Leftrightarrow4x^2=8\)

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow\)\(x=\pm\sqrt[]{2}\)

c, Ta có \(f\left(x\right)=4x^2-9\)

\(f\left(-x\right)=4\left(x\right)^2-9\)

\(=4x^2-9\) \(=f\left(x\right)\)

Vậy \(f\left(x\right)=f\left(-x\right)\)

-Chúc bạn học tốt-

a: f(-1)=-03

f(0)=-2

b: f(x)=3

=>x-2=3

hay x=5

a: f(-1)=2

f(3/2)=-21/2

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

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

bài 1:

a) y=f(0)=|1-0|+2=3

y=f(1)=|1-(-1)|+2=4

y=f(-1/2)=|1-(-1/2)|+2=7/2

b) f(x)=3 <=> |1-x|+2=3

|1-x|=3-2

|1-x|=1

=> \(\orbr{\begin{cases}1-x=1\\1-x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

f(x)=3-x <=> |1-x|+2=3-x

|1-x|=3-x-2

|1-x|=1-x

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

2.(1-x)=0

=> 1-x=0

=> x=1