\(f\left(x\right)=3\Leftrightarrow\left|x-1\right|+2=3\Leftrightarrow\left|x-1\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}x-1=1\\1-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

\(\text{1)}\)

\(\text{Thay }x=-2,\text{ ta có: }f\left(-2\right)-5f\left(-2\right)=\left(-2\right)^2\Rightarrow f\left(-2\right)=-1\)

\(\Rightarrow f\left(x\right)=x^2+5f\left(-2\right)=x^2-5\)

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

\(\text{2)}\)

\(\text{Thay }x=1,\text{ ta có: }f\left(1\right)+f\left(1\right)+f\left(1\right)=6\Rightarrow f\left(1\right)=2\)

\(\text{Thay }x=-1,\text{ ta có: }f\left(-1\right)+f\left(-1\right)+2=6\Rightarrow f\left(-1\right)=2\)

\(\text{3)}\)

\(\text{Thay }x=2,\text{ ta có: }f\left(2\right)+3f\left(\frac{1}{2}\right)=2^2\text{ (1)}\)

\(\text{Thay }x=\frac{1}{2},\text{ ta có: }f\left(\frac{1}{2}\right)+3f\left(2\right)=\left(\frac{1}{2}\right)^2\text{ (2)}\)

\(\text{(1) - 3}\times\text{(2) }\Rightarrow f\left(2\right)+3f\left(\frac{1}{2}\right)-3f\left(\frac{1}{2}\right)-9f\left(2\right)=4-\frac{1}{4}\)

\(\Rightarrow-8f\left(2\right)=\frac{15}{4}\Rightarrow f\left(2\right)=-\frac{15}{32}\)

sai 1 chút chỗ cÂU 3

nhân vs 3 thì phải là 1/12

thay x bằng ? mik cũng ko bit làm lên vào đây tham khảo hihihihi

Ta có: f(x) = 3 => y = 3

Thay vào ta có:

\(\left|3x-1\right|-2\) = 3

=> \(\Rightarrow\left|3x-1\right|=3+2=5\)

+) 3x - 1 = 5

=> 3x = 5 + 1 = 6

=> x = \(\frac{6}{3}=2\)

+) 3x - 1 = -5

=> 3x = -5 + 1 = -4

=> x = \(\frac{-4}{3}\)

Vậy x = 2 hoặc x = \(\frac{-4}{3}\)

Ta có: \(y=f\left(x\right)=\left|3x-1\right|-2\)

Khi \(f\left(x\right)=3\) thì \(3=\left|3x-1\right|-2\)

\(\Rightarrow\left|3x-1\right|=5\)

\(\Rightarrow3x-1=\pm5\)

+) \(3x-1=5\Rightarrow x=2\)

+) \(3x-1=-5\Rightarrow x=\frac{-4}{3}\)

Vậy \(x\in\left\{2;\frac{-4}{3}\right\}\)

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

\(f\left(-2\right)=2\cdot\left(-2\right)^2-5=2\cdot4-5=8-5=3\)

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

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

Thay x=3 vào hàm số \(f\left(x\right)=2x^2-5\), ta được: 

\(f\left(3\right)=2\cdot3^2-5=2\cdot9-5=18-5=13\)

Vậy: f(-2)=3

f(1)=-3

f(3)=13

b) Để f(x)=3 thì \(2x^2-5=3\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

Vậy: Để f(x)=3 thì \(x\in\left\{2;-2\right\}\)

* Tính f(0)

x = 0\(\Rightarrow f\left(0\right)+2.f\left(0\right)=1\Rightarrow3f\left(0\right)=1\Rightarrow f\left(0\right)=\frac{1}{3}\)

* Tính f(1)

x = 1 \(\Rightarrow f\left(1\right)+3.f\left(-1\right)=2\)(1)

x = -1 \(\Rightarrow f\left(-1\right)+3.f\left(1\right)=0\Rightarrow3f\left(-1\right)+9f\left(1\right)=0\)(2)

Lấy (2) - (1), ta được: \(8f\left(1\right)=-2\Rightarrow f\left(1\right)=\frac{-1}{4}\)

* Tính f(2)

x = 2 \(\Rightarrow f\left(2\right)+6.f\left(-2\right)=3\)(3)

x = -2 \(\Rightarrow f\left(-2\right)+6.f\left(2\right)=-1\Rightarrow6f\left(-2\right)+36f\left(2\right)=-6\)(4)

Lấy (4) - (3), ta được: \(35f\left(2\right)=-9\Rightarrow f\left(2\right)=\frac{-9}{35}\)