\(\dfrac{12}{x}\)

a.Tính f(5)

f(-3)

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NV
15 tháng 12 2020

\(f\left(5\right)=\dfrac{12}{5}\) ; \(f\left(-3\right)=\dfrac{12}{-3}=-4\)

10 tháng 12 2017

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

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

10 tháng 12 2017

cho hàm số y=f(x)=12x12x

Tính f(5);f(-3)

Bài làm

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

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

6 tháng 12 2017

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

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 \(x=26\)

29 tháng 11 2017

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 \(x=1\)

a: \(f\left(-1\right)=3-7=-4\)

\(f\left(\dfrac{1}{5}\right)=\dfrac{3}{25}-7=\dfrac{-172}{25}\)

b: f(x)=-20/3

\(\Leftrightarrow3x^2-7=-\dfrac{20}{3}\)

\(\Leftrightarrow3x^2=\dfrac{1}{3}\)

\(\Leftrightarrow x^2=\dfrac{1}{9}\)

=>x=1/3 hoặc x=-1/3

a) \(y=f\left(x\right)=\dfrac{6}{x}\)
*) \(f\left(1\right)=\dfrac{6}{1}=6\Rightarrow y=f\left(1\right)=6\)
*) \(f\left(1.5\right)=\dfrac{6}{1.5}=1,2\Rightarrow y=f\left(1.5\right)=1,2\)
*) \(f\left(3\right)=\dfrac{6}{3}=2\Rightarrow y=f\left(3\right)=2\)
*) \(f\left(-\dfrac{2}{3}\right)=\dfrac{6}{-\dfrac{2}{3}}=-9\Rightarrow y=f\left(-\dfrac{2}{3}\right)=-9\)
b) \(x:y=3\)
Tại \(y=-2\)
\(\Rightarrow x:\left(-2\right)=3\)
\(\Rightarrow x=3.\left(-2\right)\)
\(\Rightarrow x=-6\)
Vậy \(x=-6\)

- Xin lỗi ☹ làm lại cậu b cho ~ tại đề bài không rõ
b) \(y=f\left(x\right)=\dfrac{6}{x}\)
*)Tại y=3 \(\Rightarrow3=\dfrac{6}{x}\) \(\Rightarrow x=2\)
Vậy tại y = 3 thì x = 2
*) Tại y = -2 \(\Rightarrow-2=\dfrac{6}{x}\Rightarrow x=-3\)
Vậy tại y = -2 thì x = -3

1 tháng 8 2018

\(f\left(1\right)=-\dfrac{3}{2}.1=-\dfrac{3}{2}\)

\(f\left(-1\right)=-\dfrac{3}{2}.\left(-1\right)=\dfrac{3}{2}\)

\(f\left(2\right)=-\dfrac{3}{2}.2=-3\)

\(f\left(-2\right)=-\dfrac{3}{2}.\left(-2\right)=3\)

\(f\left(\dfrac{1}{2}\right)=-\dfrac{3}{2}.\dfrac{1}{2}=\dfrac{-3}{4}\)

\(f\left(-\dfrac{1}{2}\right)=-\dfrac{3}{2}.\left(-\dfrac{1}{2}\right)=\dfrac{3}{4}\)

\(f\left(a\right)< f\left(-a\right)\)

11 tháng 12 2017

a,

f(-1) = |1 - (-1)| + 2 = 4

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

b,

f(x) = 5

<=> |1-x| + 2 = 5

<=> |1-x| = 3

<=> \(\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)

11 tháng 11 2017

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

a) \(f\left(0\right)=2.0+\dfrac{1}{2}=0+\dfrac{1}{2}=\dfrac{1}{2}\)

b) \(f\left(\dfrac{1}{2}\right)=2.\dfrac{1}{2}+\dfrac{1}{2}=1+\dfrac{1}{2}=\dfrac{3}{2}\)

c) \(f\left(-2\right)=2.\left(-2\right)+\dfrac{1}{2}=-4+\dfrac{1}{2}=\dfrac{-7}{2}\)

12 tháng 4 2020

f(x)=2x+12f(x)=2x+12

a) f(0)=2.0+12=0+12=12f(0)=2.0+12=0+12=12

b) f(12)=2.12+12=1+12=32f(12)=2.12+12=1+12=32

c) f(−2)=2.(−2)+12=−4+12=−72

a: f(0)=2

f(-1/2)=2-1/4=7/4

b: \(f\left(x-1\right)=2-\left(x-1\right)^2=2-\left(1-x\right)^2=f\left(1-x\right)\)

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)