cho h/số y=f(x)=\(\dfrac{12}{x}\)
a.Tính f(5)
f(-3)
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f(5)\(=\)\(\dfrac{12}{5}=2,4\)
f(-3)\(=\)\(\dfrac{12}{-3}=-4\)
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
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)
\(y=f\left(x\right)=2x^2\) | -5 | -3 | 0 | 3 | 5 |
f(x) | 50 | 18 | 0 | 18 | 50 |
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 = \(3^3\) )
\(f\left(5\right)=\dfrac{12}{5}\) ; \(f\left(-3\right)=\dfrac{12}{-3}=-4\)