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a). x = -5 => y = 5.(-5)-1 = -26 => y = -26
x = -4 => y = 5.(-4)-1 = -21 => y = -21
x = -3 => y = 5.(-3)-1 = -16 => y = -16
x = -2 => y = 5.(-2)-1 = -11
x = 0 => y = 5.0-1 = -1
x = \(\frac{1}{5}\) => y = 5.\(\frac{1}{5}\)-1 = 0 => y = 0
*Vậy các giá trị tương ứng của y là: -26 ; -21 ; -16 ; -11 ; -1 ; 0.
b). y = f ( \(\frac{1}{2}\)) = 3.( \(\frac{1}{2}\))2 + 1 = \(\frac{7}{4}\)=> f (\(\frac{1}{2}\)) = \(\frac{7}{4}\)
y = f (1) = 3.12 + 1 = 4 => f (1) = 4
y = f (3) = 3.32 + 1 = 28 => f (3) = 28.
(Bài này mình tự làm. Chúc bạn học tốt môn Math ^^)
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
\(f\left(1\right)=-\dfrac{3}{2}.1=-\dfrac{3}{2}\)
\(f\left(-1\right)=-\dfrac{3}{2}.\left(-1\right)=\dfrac{3}{2}\)
\(f\left(2\right)=-\dfrac{3}{2}.2=-3\)
\(f\left(-2\right)=-\dfrac{3}{2}.\left(-2\right)=3\)
\(f\left(\dfrac{1}{2}\right)=-\dfrac{3}{2}.\dfrac{1}{2}=\dfrac{-3}{4}\)
\(f\left(-\dfrac{1}{2}\right)=-\dfrac{3}{2}.\left(-\dfrac{1}{2}\right)=\dfrac{3}{4}\)
\(f\left(a\right)< f\left(-a\right)\)
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
<=> \(\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
\(y=f\left(x\right)=\dfrac{5}{x-1}\)
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 \(x=26\)
P/s: Câu c sủa đề đi, như đề cũ không chứng minh được đâu
\(a)\) \(y=f\left(x\right)=4x^2-5\)
\(\Leftrightarrow f\left(3\right)=4.3^2-5=31\)
\(\Leftrightarrow f\left(-\frac{1}{2}\right)=4.\left(-\frac{1}{2}\right)^2-5=-4\)
\(b)\) \(f\left(x\right)=-1\)
\(\Leftrightarrow4x^2-5=-1\)
\(\Leftrightarrow4x^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)
\(c)\) Đặt \(f\left(x\right)=kx\Leftrightarrow-f\left(x\right)=-kx\)
Và \(f\left(-x\right)=k\left(-x\right)=-kx\)
Do đó chứng minh được \(-f\left(x\right)=f\left(-x\right)\)
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 \(x=1\)