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a) Ta có: \(y=f\left(x\right)=4x^2-5\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(3\right)=4.3^2-5=31\\f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2-5=-4\end{matrix}\right.\)
b) Ta có: \(f\left(x\right)=-1\)
\(\Rightarrow4x^2-5=-1\)
\(\Leftrightarrow4x^2=4\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1\right\}\) thì \(f\left(x\right)=-1\)
c) \(\forall x\in R,f\left(x\right)=f\left(-x\right)\Leftrightarrow f\left(-x\right)=4.\left(-x\right)^2-5=4x^2-5=f\left(x\right)\)
Vậy \(\forall x\in R\) thì \(f\left(x\right)=f\left(-x\right)\)
\(a.f\left(3\right)=4.3^2-5=31.\\ f\left(\dfrac{-1}{2}\right)=4.\left(\dfrac{-1}{2}\right)^2-5=-4.\)
\(b.f\left(x\right)=-1.\Rightarrow4x^2-5=-1.\\ \Leftrightarrow4x^2=4.\Leftrightarrow x^2=1.\\ \Leftrightarrow x=\pm1.\)
\(c.f\left(x\right)=f\left(-x\right).\\ \Rightarrow4x^2-5=4\left(-x\right)^2-5.\\ \Leftrightarrow4x^2-5=4x^2-5.\)
\(\Leftrightarrow0x=0\) (luôn đúng).
Vậy với mọi x ∈ R thì f (x)= f (-x).
\(y=f\left(x\right)=4x^2-9\)
a, \(f\left(-2\right)=4.\left(-2\right)^2-9\)
\(=16-9\)
\(=7\)
\(f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2-9\)
\(=4.\dfrac{1}{4}-9\)
\(=1-9\)
\(=-8\)
b, \(f\left(x\right)=-1\Rightarrow4x^2-9=-1\)
\(\Leftrightarrow4x^2=8\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow\)\(x=\pm\sqrt[]{2}\)
c, Ta có \(f\left(x\right)=4x^2-9\)
\(f\left(-x\right)=4\left(x\right)^2-9\)
\(=4x^2-9\) \(=f\left(x\right)\)
Vậy \(f\left(x\right)=f\left(-x\right)\)
-Chúc bạn học tốt-
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) Có: y = f(x) = 4x2 - 3
=> f(-2) = 4 . (-2) - 3
= -11
Vậy f(-2) = -11
b) Có: f(x) = 4x2 - 3
Mà f(x) = 1
=> 4x2 - 3 = 1
<=> 4x2 = 4
<=> x2 = 1
<=> x = 1 hoặc x = -1
Vậy x = 1 hoặc x = -1 thì f(x) = 1.
c) Có: f(x) = 4x2 - 3
Mà f(x) = x
=> 4x2 - 3 = x
<=> 4x2 - 3 - x = 0
<=> (4x2 + 3x) - (4x + 3) = 0
<=> x(4x + 3) - (4x+ 3) = 0
<=> (x - 1)(4x + 3) = 0
<=> x - 1 = 0 hoặc 4x + 3 = 0
<=> x = 1 hoặc 4x = -3
<=> x = 1 hoặc x = \(-\frac{3}{4}\)
Vậy x = 1 hoặc x = \(-\frac{3}{4}\) thì f(x) = x.
Linz
a, \(f\left(-2\right)=4\left(-2\right)^2-3=16-3=13\)
b, \(f\left(x\right)=1\)hay \(f\left(x\right)=4x^2-3=1\)
\(\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\)
c, \(f\left(x\right)=x\)hay \(4x^2-3=x\)
\(\Leftrightarrow4x^2-3-x=0\Leftrightarrow3x^2+x^2-3-x=0\)
\(\Leftrightarrow3\left(x^2-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[3\left(x+1\right)+x\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+3\right)=0\Leftrightarrow x=1;-\frac{3}{4}\)