Question

Cho \(f\left(x\right)=ax^2+bx+c\)  (a ,b,c là các số thực )

a) Biết 10a+2b-5c=0 . Chứng minh\(f\left(-1\right).f\left(-4\right)\ge0\)

b) Biết 13a + b + 2c=0 . Chứng minh \(f\left(-2\right).f\left(3\right)\le0\)

Answer 1

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)

Answer 2

a. 

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(
−
1
)
=
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−
�
+
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f(−1)=a−b+c

�
(
−
4
)
=
16
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−
4
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+
�
f(−4)=16a−4b+c

⇒
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(
−
4
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−
6
�
(
−
1
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=
16
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−
4
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+
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−
6
(
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−
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+
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=
10
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+
2
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−
5
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=
0
⇒f(−4)−6f(−1)=16a−4b+c−6(a−b+c)=10a+2b−5c=0

⇒
�
(
−
4
)
=
6
�
(
−
1
)
⇒f(−4)=6f(−1)

⇒
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(
−
1
)
�
(
−
4
)
=
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(
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1
)
.
6
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(
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1
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=
6
[
�
(
−
1
)
]
2
≥
0
⇒f(−1)f(−4)=f(−1).6f(−1)=6[f(−1)] 
2
 ≥0 (đpcm)

b.

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(
−
2
)
=
4
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−
2
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+
�
f(−2)=4a−2b+c

�
(
3
)
=
9
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+
3
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+
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f(3)=9a+3b+c

⇒
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−
2
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+
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(
3
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=
13
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+
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+
2
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=
0
⇒f(−2)+f(3)=13a+b+2c=0

⇒
�
(
−
2
)
=
−
�
(
3
)
⇒f(−2)=−f(3)

⇒
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(
−
2
)
�
(
3
)
=
−
[
�
(
3
)
]
2
≤
0
⇒f(−2)f(3)=−[f(3)] 
2
 ≤0 (đpcm