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 \(x=\sqrt{2}-1\)

a)  Cho hàm số y = f(x) = -2x + 3.

Ta có: f(-2)= -2.(-2)+3

                 = 4+3=7

Ta có: f(0)= -2.0+3

                = 0+3=3

Ta có: f(

Lời giải:

a.

$f(-2)=(-2)(-2)+3=7$

$f(0)=(-2).0+3=3$

$f(\frac{-1}{2})=(-2).\frac{-1}{2}+3=4$

b.

$f(x)=-2x+3=5$

$\Rightarrow -2x=2$

$\Rightarrow x=-1$

$f(x)=-2x+3=1$

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

$\Rightarrow x=1$

a) * f(-2)

=-2.(-2)+1

=2

 * f(3)

=-2.3+1

=-5

b) hàm số y=-2x+1

 với x=-1 thì y=3 không bằng 1 

Vậy M(-1,1)ko thuộc đồ thị hàm số f(x)

c) ta có 1>0 

=> -2x+1=1

      -2x=1-1

      -2x=0

       x=0/(-2)

       x=0

=> x=0

vậy x=0 thì f(x)>0

nhớ k giùm mình nha

a)\(F\left(-2\right)=-2.\left(-2\right)+1=5\)

    \(F\left(\frac{1}{2}\right)=-2.\left(\frac{1}{2}\right)+1=0\)

    \(F\left(3\right)=-2.3+1=-5\)

    \(F\left(1\right)=-2.1+1=-1\)

a)f(-2)=-2.(-2)+1=4

f(1/2)=-2.(1/2)+1=0

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

b)M(-1;1)=>x=-1;y=1

Thayx=-1vao ham so y=-2x+1, ta duoc:

y=-2.(-1)+1=3khac1

vay M(-1;1)khong thuoc do thi ham so y=-2x+1

c)x<0 thi f(x)>0

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

a) Thay x=-2 vào hàm số f(x)=|3x-1|, ta được:

\(f\left(-2\right)=\left|3\cdot\left(-2\right)-1\right|=\left|-6-1\right|=7\)

Thay x=2 vào hàm số \(f\left(x\right)=\left|3x-1\right|\), ta được:

\(f\left(2\right)=\left|3\cdot2-1\right|=\left|6-1\right|=5\)

Thay \(x=-\dfrac{1}{4}\) vào hàm số \(f\left(x\right)=\left|3x-1\right|\), ta được:

\(f\left(-\dfrac{1}{4}\right)=\left|3\cdot\dfrac{-1}{4}-1\right|=\left|-\dfrac{3}{4}-\dfrac{4}{4}\right|=\dfrac{7}{4}\)

Thay \(x=\dfrac{1}{4}\) vào hàm số \(f\left(x\right)=\left|3x-1\right|\), ta được:

\(f\left(\dfrac{1}{4}\right)=\left|3\cdot\dfrac{1}{4}-1\right|=\left|\dfrac{3}{4}-1\right|=\dfrac{1}{4}\)

Vậy: f(-2)=7; f(2)=5; \(f\cdot\left(-\dfrac{1}{4}\right)=\dfrac{7}{4}\); \(f\left(\dfrac{1}{4}\right)=\dfrac{1}{4}\)

b) Để f(x)=10 thì \(\left|3x-1\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=10\\3x-1=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=11\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=-3\end{matrix}\right.\)

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

mà \(\left|3x-1\right|\ge0\forall x\)

nên \(x\in\varnothing\)

Ta có : \(f\left(x\right)=2x^2+3\)

b, \(f\left(x\right)=11\)hay \(2x^2+3=11\)

\(\Leftrightarrow2x^2=8\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

Vậy x = 2 ; -2 khi f(x) = 11

a ) Khi \(f\left(2\right)\)

\(5.2+1-\left|5.2-3\right|\)

\(=10+1-\left|10-3\right|\)

\(=10+1-7\)

\(=4\)

Khi \(f\left(-7\right)\)

\(5.\left(-7\right)+1-\left|2.\left(-7\right)-3\right|\)

\(=-35+1-\left|-14-3\right|\)

\(=-34-\left|-17\right|\)

\(=-34-17\)

\(=-51\)

b ) Khi \(f\left(2\right)\) , thì :

\(5.2+1-2.2-3\)

\(=10+1-4-3\)

\(=4\)

Khi \(f\left(-7\right)\) , thì :

\(5.\left(-7\right)+1-2.\left(-7\right)-3\)

\(=-35+1+14-3\)

.\(=-23\)

\(a.\)

Ta có : \(y=f\left(x\right)=5x+1-\left|2x-3\right|\)

\(\Rightarrow f\left(2\right)=5.\left(2\right)+1-\left|2.\left(2\right)-3\right|=10\)

\(\Rightarrow f\left(x\right)=5.\left(-7\right)+1-\left|2.\left(-7\right)-3\right|=-51\)

\(b.\)

Ta có : \(y=f\left(x\right)=5x+1-2x-3\)

\(\Rightarrow f\left(2\right)=5.\left(2\right)+1-2.\left(2\right)-3=4\)

\(f\left(-7\right)=5.\left(-7\right)+1-2.\left(-7\right)-3=-23\)

a: f(-3)=10

f(0)=-8

f(1)=-6

f(2)=0

b: f(x)=0

=>(x-2)(x+2)=0

=>x=2 hoặc x=-2