\(y=f\left(x\right)=2x^2-1.\)

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

\(f\left(-2\right)=2.\left(-2\right)^2-1=7.\)

\(f\left(1\right)=2.1^2-1=1.\)

\(f\left(2\right)=2.2^2-1=7.\)

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:

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

\(f\left(\dfrac{1}{2}\right)=-\dfrac{1}{2}\)

\(f\left(0\right)=-1\)

\(f\left(-5\right)=49\)

Bài 2:

Bạn lấy toạ độ điểm A(1;3); B(-1;-3), C(0;0). Đồ thị y=3x một đường thẳng đi qua gốc toạ độ O nhé!

Ta có: f(1) = 2.12 – 5 = 2 – 5 = -3

f(-2) = 2.(-2)2 – 5 = 8 – 5 = 3

f(0) = 2.02 – 5 = 0 – 5 = -5

f(2) = 2.22 – 5 = 2.4 – 5 = 3

a/ Thay x =0 vào hàm số f(x) = 2x² - 10 ta có

f(0) = 2 . 0 - 10 = -10

Thay x = 1 vào hàm số f(x) = 2x² - 10 ta có

f(1) = 2 . 1² - 10 = 2 - 10 = -8

Thay \(x=-1\dfrac{1}{2}=-\dfrac{3}{2}\)vào hàm số f(x) ta có

\(f\left(-1\dfrac{1}{2}\right)=2.\left(-\dfrac{3}{2}\right)^2-10=\dfrac{9}{2}-\dfrac{20}{2}=-\dfrac{11}{2}\)

b/ f(x) = -2

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

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

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-

\(C_1:f\left(1\right)=2-5=-3\\ f\left(2\right)=2\cdot4-5=3\\ f\left(0\right)=0-5=-5\\ f\left(-1\right)=2-5=-3\\ f\left(-2\right)=2\cdot4-5=3\\ C_2:\begin{matrix}x&1&2&0&-1&-2\\y=2x^2-5&-3&3&-5&-3&3\end{matrix}\)

(1)

a) x=\(\dfrac{-1}{12}-\dfrac{2}{3}\)=\(\dfrac{-3}{4}\)

b) 2x+1=3 => 2x=3-1=2 => x=1

(2)

f(2)=2.2²+4=12

f(-1)=2.(-1)²+4=6

(1)

a) \(x+\dfrac{2}{3}=-\dfrac{1}{12}\\ \Rightarrow x=-\dfrac{1}{12}-\dfrac{2}{3}\\ \Rightarrow x=\dfrac{-1}{12}-\dfrac{8}{12}\\ \Rightarrow x=-\dfrac{9}{12}=-\dfrac{3}{4}\)

Vậy \(x=-\dfrac{3}{4}\)

b) \(\left(2x+1\right)^2=9\\ \Rightarrow\left(2x+1\right)^2=3^2=\left(-3\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=2\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{-2;1\right\}\)

(2)

\(y=f\left(x\right)=2x^2+4\\ f\left(2\right)=2\cdot2^2+4=8+4=12\\ f\left(-1\right)=2\cdot\left(-1\right)^2+4=2+4=6\)

Vậy \(f\left(2\right)=12\\ f\left(-1\right)=6\)