\(f\left(3\right)=3a-3=9\)

\(3a=12\Rightarrow a=4\)

\(f\left(5\right)=5a-3=11\)

\(5a=14\Rightarrow a=\dfrac{14}{5}\)

\(f\left(-1\right)=-a-3=6\)

\(-a=9\Rightarrow a=9\)

a) \(f\left(x\right)=2x^2+5x-3\)

\(f\left(1\right)=2.1^2+5.1-3=2+5-3=4\)

\(f\left(0\right)=-3\)

\(f\left(1,5\right)=2.\left(1,5\right)^2+5.1,5-3=2.2,25+7,5-3=9\)

Bài 1:

\(f\left(x\right)=5x-3.\)

+ \(f\left(x\right)=0\)

\(\Rightarrow5x-3=0\)

\(\Rightarrow5x=0+3\)

\(\Rightarrow5x=3\)

\(\Rightarrow x=3:5\)

\(\Rightarrow x=\frac{3}{5}\)

Vậy \(x=\frac{3}{5}.\)

+ \(f\left(x\right)=1\)

\(\Rightarrow5x-3=1\)

\(\Rightarrow5x=1+3\)

\(\Rightarrow5x=4\)

\(\Rightarrow x=4:5\)

\(\Rightarrow x=\frac{4}{5}\)

Vậy \(x=\frac{4}{5}.\)

+ \(f\left(x\right)=-2010\)

\(\Rightarrow5x-3=-2010\)

\(\Rightarrow5x=\left(-2010\right)+3\)

\(\Rightarrow5x=-2007\)

\(\Rightarrow x=\left(-2007\right):5\)

\(\Rightarrow x=-\frac{2007}{5}\)

Vậy \(x=-\frac{2007}{5}.\)

Làm tương tự với \(f\left(x\right)=2011.\)

Chúc bạn học tốt!

a)\(f\left(1\right)=2.1^2+5.1-3=2+5-3=4\)

\(f\left(0\right)=0+0-3=-3\)

\(f\left(1,5\right)=2.\left(1,5\right)^2-5.1,5-3=4,5-7,5-3=-6\)

b)\(f\left(3\right)=3a-3=9=>>3a=12=>a=4\)

\(f\left(5\right)=5a-3=11=>5a=14=>a=\dfrac{14}{5}\)

\(f\left(-1\right)=-a-3=6=>-a=9=>a=-9\)

`a)`

`@f(1)=2.1^2+5.1-3=2.1+5-3=2+5-3=4`

`@f(0)=2.0^2+5.0-3=-3`

`@f(1,5)=2.(1,5)^2+5.1,5-3=4,5+7,5-3=9`

_____________________________________________________

`b)`

`***f(3)=9`

`=>3a-3=9`

`=>3a=12=>a=4`

`***f(5)=11`

`=>5a-3=11`

`=>5a=14=>a=14/5`

`***f(-1)=6`

`=>-a-3=6`

`=>-a=9=>a=-9`

a: f(1)=2+5-3=4

f(0)=-3

f(1,5)=4,5+7,5-3=9

b: f(3)=9 nên 3a-3=9

hay a=4

f(5)=11 nên 5a-3=11

hay a=14/5

f(-1)=6 nên -a-3=6

=>-a=9

hay a=-9

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 $-1=5-2x\Rightarrow x=3$

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 \(x=\sqrt{2}-1\)