f(5)\(=\)\(\dfrac{12}{5}=2,4\)

f(-3)\(=\)\(\dfrac{12}{-3}=-4\)

cho hàm số y=f(x)=

Tính f(5);f(-3)

Bài làm

f(5) = 12 : 5 = 2,4

f(-3) = 12 : (-3) = -4

f(5)=12/5=2,5

f(-3)=12/-3=-4

a)

f(-2) = 4 . (-2) - 3 = -8 - 3 = -11

f(3) = 4 . 3 - 3 = 12 - 3 = 9

f(2) = 4 . 2 - 3 = 8 - 3 = 5

b)

f(x) = 1

=> 4x - 3 = 1

=> 4x = 1 + 3

=> 4x = 4

=> x = 4 / 4

=> x = 1

Vậy x = 1 thì f(x) = 1

\(y=f\left(x\right)=\dfrac{5}{x-1}\)

a) ĐKXĐ khi: \(x-1\ne0\) \(\Leftrightarrow x\ne1\)

b)

\(y=f\left(-2\right)=\dfrac{5}{-2-1}=\dfrac{5}{-\left(2+1\right)}=\dfrac{5}{-3}=\dfrac{-5}{3}\)

\(y=f\left(\dfrac{1}{3}\right)=\dfrac{5}{\dfrac{1}{3}-1}=\dfrac{5}{\dfrac{1}{3}-\dfrac{3}{3}}=\dfrac{5}{\dfrac{-2}{3}}=\dfrac{-15}{2}\)

c)

* \(y=f\left(x\right)=\dfrac{5}{x-1}=-1\)

\(\Rightarrow x-1=5:\left(-1\right)\)

\(\Rightarrow x-1=-5\)

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

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

\(\Rightarrow x=-4\)

Vậy \(y=-1\) thì \(x=-4\)

* \(y=f\left(x\right)=\dfrac{5}{x-1}=1\)

\(\Rightarrow x-1=5:1\)

\(\Rightarrow x-1=5\)

\(\Rightarrow x=5+1\)

\(\Rightarrow x=6\)

Vậy \(y=1\) thì \(x=6\)

* \(y=f\left(x\right)=\dfrac{5}{x-1}=\dfrac{1}{5}\)

\(\Rightarrow x-1=5:\dfrac{1}{5}\)

\(\Rightarrow x-1=5.5\)

\(\Rightarrow x-1=25\)

\(\Rightarrow x=25+1\)

\(\Rightarrow x=26\)

Vậy \(y=\dfrac{1}{5}\) thì \(x=26\)

a) \(y=f\left(x\right)=1-5x\)

\(y=f\left(1\right)=1-5.1=1-5=-4\)

\(y=f\left(-2\right)=1-5.\left(-2\right)=1-\left(-10\right)=1+10=11\)

\(y=f\left(\dfrac{1}{5}\right)=1-5.\dfrac{1}{5}=1-1=0\)

\(y=f\left(\dfrac{-3}{5}\right)=1-5.\left(\dfrac{-3}{5}\right)=1-\left(-3\right)=1+3=4\)

b) \(y=f\left(x\right)=1-5x=-4\)

\(\Rightarrow5x=1-\left(-4\right)\)

\(\Rightarrow5x=1+4\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=\dfrac{5}{5}=1\)

Vậy \(f\left(x\right)=-4\) thì \(x=1\)

Giải:

Có \(y=f\left(x\right)=\dfrac{1}{2}x+5\)

\(\Leftrightarrow f\left(0\right)=\dfrac{1}{2}.0+5=0+5=5\)

\(\Leftrightarrow f\left(1\right)=\dfrac{1}{2}.1+5=\dfrac{1}{2}+5=\dfrac{11}{2}\)

\(\Leftrightarrow f\left(2\right)=\dfrac{1}{2}.2+5=1+5=6\)

\(\Leftrightarrow f\left(3\right)=\dfrac{1}{2}.3+5=\dfrac{3}{2}+5=\dfrac{13}{2}\)

\(\Leftrightarrow f\left(-2\right)=\dfrac{1}{2}.\left(-2\right)+5=-1+5=4\)

\(\Leftrightarrow f\left(-10\right)=\dfrac{1}{2}.\left(-10\right)+5=-5+5=0\)

Vậy ...

Chúc bạn học tốt!

\(y=f\left(x\right)=\dfrac{1}{2}x+5\)

Ta có : \(f\left(0\right)=\dfrac{1}{2}.0+5=5\)

\(f\left(1\right)=\dfrac{1}{2}.1+5=\dfrac{11}{2}=5,5\)

\(f\left(2\right)=\dfrac{1}{2}.2+5=6\)

\(f\left(3\right)=\dfrac{1}{2}.3+5=\dfrac{13}{2}=6,5\)

\(f\left(-2\right)=\dfrac{1}{2}.\left(-2\right)+5=4\)

\(f\left(-10\right)=\dfrac{1}{2}.\left(-10\right)+5=0\)

#HỌC TỐT ~~~

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 \(x=\sqrt{2}-1\)

Bài 3:

a: f(-1)=-2

f(1/2)=1

b: f(x)=5

=>2x=5

=>x=5/2

c: f(5a)=2*5a=10a

5*f(a)=5*2a=10a

=>f(5a)=5*f(a)

a,\(f\left(-4\right)=2.\left(-4\right)^3-3.\left(-4\right)=2.\left(-64\right)+12=-128+12=-116\)

\(f\left(-2\right)=2.\left(-2\right)^3-3.\left(-2\right)=2.\left(-8\right)+6=-16+6=-10\)

\(f\left(0\right)=2.0^3-3.0=2.0-0=0-0=0\)

\(f\left(\dfrac{2}{3}\right)=2.\left(\dfrac{2}{3}\right)^3-3.\left(\dfrac{2}{3}\right)=2.\dfrac{8}{27}-2=\dfrac{16}{27}-2=\dfrac{-38}{27}\)

b,

\(f\left(x\right)=25\rightarrow y=25\)

Ta có : \(x^3-2=25\)

\(\rightarrow x^3=27\)

\(\Rightarrow x=3\) ( Vì 27 = \(3^3\) )