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Lời giải:
a.
$f(-1)=a-b+c$
$f(-4)=16a-4b+c$
$\Rightarrow f(-4)-6f(-1)=16a-4b+c-6(a-b+c)=10a+2b-5c=0$
$\Rightarrow f(-4)=6f(-1)$
$\Rightarrow f(-1)f(-4)=f(-1).6f(-1)=6[f(-1)]^2\geq 0$ (đpcm)
b.
$f(-2)=4a-2b+c$
$f(3)=9a+3b+c$
$\Rightarrow f(-2)+f(3)=13a+b+2c=0$
$\Rightarrow f(-2)=-f(3)$
$\Rightarrow f(-2)f(3)=-[f(3)]^2\leq 0$ (đpcm)
a.
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f(−4)=16a−4b+c
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⇒f(−4)−6f(−1)=16a−4b+c−6(a−b+c)=10a+2b−5c=0
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⇒f(−4)=6f(−1)
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⇒f(−1)f(−4)=f(−1).6f(−1)=6[f(−1)]
2
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b.
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f(−2)=4a−2b+c
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f(3)=9a+3b+c
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⇒f(−2)+f(3)=13a+b+2c=0
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⇒f(−2)=−f(3)
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⇒f(−2)f(3)=−[f(3)]
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≤0 (đpcm
Ta có:
\(f\left(x\right)=ax^3+bx^2+cx+d\\ f\left(x\right)=0x^3+0x^2+0x+0\)
\(\Rightarrow a=b=c=d\left(theo.pp.đa.thức.đồng.nhất\right)\\ Chúc.bạn.học.Toán.tốt.\)
\(f\left(x_1\right)=ax_1\) ; \(f\left(x_2\right)=ax_2\) ; \(f\left(x_1x_2\right)=ax_1x_2\)
Để \(f\left(x_1\right)f\left(x_2\right)=f\left(x_1x_2\right)\)
\(\Leftrightarrow ax_1.ax_2=ax_1x_2\)
\(\Leftrightarrow a^2x_1x_2=ax_1x_2\)
\(\Leftrightarrow a^2=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)
Vậy \(a=1\)
a: f(1)=1
=>\(a\cdot1^2+b\cdot1+1=1\)
=>a+b=0
f(-1)=3
=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+1=3\)
=>a-b=2
mà a+b=0
nên \(a=\dfrac{2+0}{2}=1;b=2-1=1\)
b: a=1 và b=1 nên \(f\left(x\right)=x^2+x+1\)
\(\Leftrightarrow\dfrac{n}{f\left(n\right)}=\dfrac{n}{n^2+n+1}\)
Gọi d=ƯCLN(n^2+n+1;n)
=>\(\left\{{}\begin{matrix}n^2+n+1⋮d\\n⋮d\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}n^2+n+1⋮d\\n\left(n+1\right)⋮d\end{matrix}\right.\)
=>\(\left(n^2+n+1\right)-n\left(n+1\right)⋮d\)
=>\(1⋮d\)
=>d=1
=>ƯCLN(n^2+n+1;n)=1
=>\(\dfrac{n}{f\left(n\right)}=\dfrac{n}{n^2+n+1}\) là phân số tối giản
f(0) = 1
\(\Rightarrow\) a.02 + b.0 + c = 1
\(\Rightarrow\) c = 1
Vậy hệ số a = 0; b = 0; c = 1
f(1) = 2
\(\Rightarrow\) a.12 + b.1 + c = 2
\(\Rightarrow\) a + b + c = 2
Vậy hệ số a = 1; b = 1; c = 1
f(2) = 4
\(\Rightarrow\) a.22 + b.2 + c = 4
\(\Rightarrow\) 4a + 2b + c = 4
Vậy hệ số a = 4; b = 2; c = 1
Chúc bn học tốt! (chắc vậy :D)
1.a) Theo đề bài,ta có: \(f\left(-1\right)=1\Rightarrow-a+b=1\)
và \(f\left(1\right)=-1\Rightarrow a+b=-1\)
Cộng theo vế suy ra: \(2b=0\Rightarrow b=0\)
Khi đó: \(f\left(-1\right)=1=-a\Rightarrow a=-1\)
Suy ra \(ax+b=-x+b\)
Vậy ...
\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)
\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)
Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)
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 \(f\left(x\right)\ne0\forall x\)