\(f\left(-1\right)=3\cdot\left(-1\right)^2-1=2\\ f\left(1\right)=3\cdot1^2-1=2\\ f\left(2\right)=3\cdot2^2-1=11\\ f\left(\dfrac{5}{2}\right)=3\cdot\left(\dfrac{5}{2}\right)^2-1=3\cdot\dfrac{25}{4}-1=\dfrac{71}{4}\)

\(f\left(-1\right)=3\cdot1-1=2\)

\(f\left(1\right)=3\cdot1^2-1=2\)

\(F\left(2\right)=3\cdot2^2-1=11\)

\(f\left(\dfrac{5}{2}\right)=3\cdot\dfrac{25}{4}-1=\dfrac{71}{4}\)

Bài 1: 

Thay x=1 vào hàm số \(y=f\left(x\right)=2x^2-5\), ta được:

\(f\left(1\right)=2\cdot1^2-5=2-5=-3\)

Thay x=-2 vào hàm số \(y=f\left(x\right)=2x^2-5\), ta được:

\(f\left(-2\right)=2\cdot\left(-2\right)^2-5=2\cdot4-5=3\)

Thay x=0 vào hàm số \(y=f\left(x\right)=2x^2-5\), ta được: 

\(f\left(0\right)=2\cdot0^2-5=-5\)

Thay x=2 vào hàm số \(y=f\left(x\right)=2x^2-5\), ta được:

\(f\left(2\right)=2\cdot2^2-5=8-5=3\)

Thay \(x=\dfrac{1}{2}\) vào hàm số \(y=f\left(x\right)=2x^2-5\), ta được:

\(f\left(\dfrac{1}{2}\right)=2\cdot\left(\dfrac{1}{2}\right)^2-5=2\cdot\dfrac{1}{4}-5=-\dfrac{9}{2}\)

Vậy: f(1)=-3; f(-2)=3; f(0)=-5; f(2)=3; \(f\left(\dfrac{1}{2}\right)=-\dfrac{9}{2}\)

Bài 1:

\(f(x)=2x^2-5\) thì:

$f(1)=2.1^2-5=-3$

$f(-2)=2(-2)^2-5=3$

$f(0)=2.0^2-5=-5$

$f(2)=2.2^2-5=3$

$f(\frac{1}{2})=2(\frac{1}{2})^2-5=\frac{-9}{2}$

Bài 2:

a) $f(x)=5-2x$ thì:

$f(-2)=5-2(-2)=9$

$f(-1)=5-2(-1)=7$

$f(0)=5-2.0=5$

$f(3)=5-2.3=-1$

b) Với $y=5$ thì $5=5-2x\Rightarrow x=0$

Với $y=3$ thì $3=5-2x\Rightarrow x=1$

Với $y=-1$ thì $-1=5-2x\Rightarrow x=3$

a: f(0)=1

\(f\left(-\dfrac{1}{3}\right)=1-3\cdot\left(-\dfrac{1}{3}\right)^2=1-3\cdot\dfrac{1}{9}=1-\dfrac{1}{3}=\dfrac{2}{3}\)

a: f(0)=1

\(f\left(-\dfrac{1}{3}\right)=1-3\cdot\left(-\dfrac{1}{3}\right)^2=1-3\cdot\dfrac{1}{9}=1-\dfrac{1}{3}=\dfrac{2}{3}\)

a: f(5)=29

f(2)=4-2=2

f(1)=1-2=-1

f(0)=-2

\(f\left(2\right)=2^2-2=2\)

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

\(f\left(0\right)=0^2-2=-2\)

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

\(f\left(-2\right)=\left(-2\right)^2-2=2\)

giúp tớ đi =))

\(f\left(x\right)=3x^2+1\)

\(f\left(x+1\right)=3\left(x+1\right)^2+1\\ f\left(x+1\right)=3\left(x^2+2x+1\right)+1\\ f\left(x+1\right)=3x^2+6x+3+1\\ f\left(x+1\right)=3x^2+6x+4\\ f\left(x+1\right)-f\left(x\right)=3x^2+6x+4-3x^2-1\\ f\left(x+1\right)-f\left(x\right)=6x+3\)

Vậy y = f (x+1) - f (x) là hàm số bậc nhất.

Ta có : \(f\left(x\right)=x^2+2x+1=\left(x+1\right)^2\)

a, \(f\left(1\right)=\left(1+1\right)^2=2^2=4\)

 \(f\left(0\right)=\left(0+1\right)^2=1^2=1\)

b, Ta có : y = 1 hay \(f\left(x\right)=1\)

\(\Leftrightarrow\left(x+1\right)^2=1\Leftrightarrow x^2+2x+1=1\)

\(\Leftrightarrow x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)

Vậy \(x=\left\{0;-2\right\}\)

\(f\left(1\right)=3\cdot1^2+1=4\)