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a) \(f\left(3\right)=4\times3^2-5=31\)
\(f\left(-\frac{1}{2}\right)=4\times\left(-\frac{1}{2}\right)^2-5=-4\)
b) để f(x)=-1
<=>\(4x^2-5=-1\)
<=>\(4x^2=4\)
<=>\(x^2=1\)
<=>\(x=\orbr{\begin{cases}1\\-1\end{cases}}\)
Cho hàm số y = f(x) = 4x^2 +4y=f(x)=4x2+4. Tính f(-2)f(−2) ; f(2)f(2) ; f(4)f(4).
Đáp số:
f(-2) =f(−2)=
f(2) =f(2)=
f(4) =f(4)=
\(f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)
\(f\left(3\right)=a.3^2+b.3+c=9a+3b+c\)
\(f\left(-2\right)+f\left(3\right)=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\Rightarrow f\left(-2\right).f\left(3\right)\le0\)
\(f\left(-1\right)=-a+b-c+d=2\)
\(f\left(0\right)=d=1\)
\(f\left(\frac{1}{2}\right)=\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c+d=3\)
\(f\left(1\right)=a+b+c+d=7\)
Suy ra \(\hept{\begin{cases}-a+b-c=1\\\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c=2\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}2b=7\\\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c=2\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{7}{2}\\c=\frac{13}{6}\end{cases}}\)
a) \(\hept{\begin{cases}f\left(2\right)=156\\f\left(-3\right)=156\\f\left(-1\right)=132\end{cases}\Rightarrow\hept{\begin{cases}4a+2b+c=156\\9a-3b+c=156\\a-b+c=132\end{cases}\Rightarrow}\hept{\begin{cases}4a+2b+132-a+b=156\\9a-3b+132-a+b=156\\c=132-a+b\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}3a+3b=24\\8a-2b=24\\c=132-a+b\end{cases}\Rightarrow\hept{\begin{cases}a+b=8\\-4a+b=-12\\c=132-a+b\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}5a=20\\b=8-a\\c=132-a+b\end{cases}\Rightarrow\hept{\begin{cases}a=4\\b=4\\c=132\end{cases}}}\)
b) \(f\left(x\right)=4x^2+4x+132=4x^2+2x+2x+1+131=2x\left(2x+1\right)+\left(2x+1\right)+131\)
\(=\left(2x+1\right)^2+131\)
\(\left(2x+1\right)^2\ge0\forall x\Rightarrow f\left(x\right)\ge131\forall x\). Vậy \(f\left(x\right)\ne0\forall x\)
\(\left\{{}\begin{matrix}f\left(0\right)=5\Rightarrow0+0+5\Rightarrow c=5\\f\left(1\right)=0\Rightarrow a+b+5=0\\f\left(5\right)=0\Rightarrow25a+5b+5=0\end{matrix}\right.\) \(\left\{{}\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\right.\)
tu (3) => b =-1-5a
tu (2) => a-1-5a+5 =0 => a =1 ;b =-6
y =x^2 -6x +5
y(-1) =1 +6 +5 khac 3 => loai
y(-1/2) =1/4 -6/2 +5 =1/4 +2 = 9/4 nhan
Q(1/2;9/4) thuoc dths
Ta có : f(x) = ax3 + 4x(x2-x) - 4x + 8
= ax3 + 4x3 - 4x2 - 4x + 11 - 3
= x3 (a + 4) - 4x(x + 1) + 11-3
f(x) = g (x) \(\Leftrightarrow\) x3 (a + 4) - 4x(x + 1) +11-3 = x3 - 4x(bx + 1) + c-3
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a+4=1\\x+1=bx+1\\c=11\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=-3\\b=1\\c=11\end{matrix}\right.\)
vậy a = -3 , b = 1 và c = 11
f(x)=g(x)
<=>(a+4)x3-4x2-4x+8=x3-4bx2-4x+c-3
Đồng nhất thức ta được
a+4=1 a=-3
-4=-4b <=> b=1
8=c-3 c=11
Ta có: \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(0\right)=c⋮3\Rightarrow c⋮3\)
\(\left\{{}\begin{matrix}f\left(1\right)=a+b+c⋮3\\f\left(-1\right)=a-b+c⋮3\end{matrix}\right.\)
Mà \(c⋮5\)
\(\Rightarrow\left\{{}\begin{matrix}a+b⋮3\\a-b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a⋮3\\2b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a⋮3\\b⋮3\end{matrix}\right.\) ( do \(\left(2;3\right)=1\) )
Vậy \(a,b,c⋮3\)