K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

3 tháng 2 2020

\(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{4}a+\frac{1}{2}b+c\)

\(\Rightarrow f\left(-2\right)=4a-2b+c\)

\(\Rightarrow f\left(\frac{1}{2}\right)+f\left(-2\right)=\frac{17}{4}a-\frac{3}{2}b+2c\)

\(\Rightarrow4\left[f\left(\frac{1}{2}\right)+f\left(-2\right)\right]=17a-6b+8c=0\)( vì 17a-6b+8c=0)

\(\Rightarrow f\left(\frac{1}{2}\right)+f\left(-2\right)=0\)

\(\Rightarrow f\left(\frac{1}{2}\right)=-f\left(-2\right)\)

\(\Rightarrow f\left(\frac{1}{2}\right).f\left(-2\right)=-\left[f\left(-2\right)\right]^2\le0\left(đpcm\right)\)

30 tháng 12 2019

tham khảo thôi nhé ko giống y sì đâu

https://olm.vn/hoi-dap/detail/213882782299.html

AH
Akai Haruma
Giáo viên
12 tháng 2 2023

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)

2 tháng 3 2023

a. 


(

1
)
=



+

f(−1)=a−b+c


(

4
)
=
16


4

+

f(−4)=16a−4b+c



(

4
)

6

(

1
)
=
16


4

+


6
(



+

)
=
10

+
2


5

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



(

4
)
=
6

(

1
)
⇒f(−4)=6f(−1)



(

1
)

(

4
)
=

(

1
)
.
6

(

1
)
=
6
[

(

1
)
]
2

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

b.


(

2
)
=
4


2

+

f(−2)=4a−2b+c


(
3
)
=
9

+
3

+

f(3)=9a+3b+c



(

2
)
+

(
3
)
=
13

+

+
2

=
0
⇒f(−2)+f(3)=13a+b+2c=0



(

2
)
=


(
3
)
⇒f(−2)=−f(3)



(

2
)

(
3
)
=

[

(
3
)
]
2

0
⇒f(−2)f(3)=−[f(3)] 
2
 ≤0 (đpcm

4 tháng 2 2021

\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)

\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)

Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)

22 tháng 2 2020

\(f\left(-1\right)=a\left(-1\right)^2+b.\left(-1\right)+c\)

\(=a-b+c\)

\(f\left(2\right)=a.2^2+b.2+c\)

\(=4a+2b+c\)

\(\Rightarrow f\left(2\right)-2.f\left(-1\right)=\left(4a+2b+c\right)-2\left(a-b+c\right)\)

\(=2a+4b-c=0\)

\(\Rightarrow f\left(2\right)=2.f\left(-1\right)\)

\(\Rightarrow f\left(2\right)\)và \(2.f\left(-1\right)\)cùng dấu

\(\Rightarrow f\left(2\right)\)và \(f\left(-1\right)\)cùng dấu

\(\Rightarrow f\left(2\right).f\left(-1\right)\ge0\)(đpcm)

22 tháng 2 2020

Ta có :\(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c\)

               \(f\left(2\right)=a.2^2+b.2+c=4a+2b+c\)

\(\implies\) \(f\left(2\right)-2f\left(-1\right)=\left(4a+2b+c\right)-2.\left(a-b+c\right)\)

\(\implies\)  \(f\left(2\right)=2.f\left(-1\right)\)

\(\implies\)  \(f\left(-1\right).f\left(2\right)=f\left(-1\right).2f\left(-1\right)=f\left(-1\right)^2.2\) \(\geq\) \(0\)

\(\implies\)  \(f\left(-1\right).f\left(2\right)\) \(\geq\)  \(0\) \(\left(đpcm\right)\)

12 tháng 3 2020

\(f\left(x\right)=ax^2+bx+c\)

\(f\left(2\right)=4a+2b+c\)

\(f\left(-1\right)=a-b+c\)

\(\Rightarrow f\left(2\right)+f\left(-1\right)=4a+2b+c+a-b+c\)

\(\Leftrightarrow f\left(2\right)+f\left(-1\right)=5a+b+2c=0\)

\(\Rightarrow f\left(2\right)+f\left(-1\right)=0\Leftrightarrow f\left(2\right)=-f\left(-1\right)\)

\(\Leftrightarrow f\left(2\right).f\left(-1\right)=-f\left(-1\right).f\left(-1\right)\le0\)

\(\Rightarrowđpcm\)

12 tháng 3 2020

thanks