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Ta có
Thay x = 1/2 : \(f\left(\frac{1}{2}\right)+3f\left(2\right)=\frac{1}{4}\)
Thay x = 2: \(f\left(2\right)+3f\left(\frac{1}{2}\right)=4\)
\(\Rightarrow\left[f\left(2\right)+3f\left(\frac{1}{2}\right)\right]-3\left[f\left(\frac{1}{2}\right)+3f\left(2\right)\right]=4-\frac{3}{4}\)
\(\Rightarrow-5f\left(2\right)=\frac{13}{4}\Leftrightarrow f\left(2\right)=-\frac{13}{20}\)
Ta có :
Thay x = 1/2 : ƒ (12 )+3ƒ (2)=14
Thay x = 2: ƒ (2)+3ƒ (12 )=4
⇒[ƒ (2)+3ƒ (12 )]−3[ƒ (12 )+3ƒ (2)]=4−34
f(2)+3f(1/2)=4
f(1/2)+3f(2)=1/4
=>f(2)+3f(1/2)=4
9f(2)+3f(1/2)=3/4
=>-8f(2)=4-3/4=13/4
=>f(2)=-13/32
Ta có 2f(x)-x.f(1/x)=x^2
Với x=2 => 2f(2)-2.f(1/2)=4 (1)
Với x=1/2 => 2 . f(1/2)- 1/2 f(2) = (1/2)^2
=> 2 .f(1/2) -1/2f(2)=1/4(2)
lấy (2)+(1) ta được 3/2 f(2)=17/4 => f(2)=17/6
Tính f(1/3) làm tương tự thay x=3 và 1/3
T ic k nha
a)Cho y=f(x)=-5x
C/m f(\(x_1\)+\(4x_2\))=f(\(x_1\))+4f(x2)
b)Cho f(x)+3f(\(\frac{1}{2}\))=x2.Tính f(2)
thay x=2 và x=1/2 ta có
\(\hept{\begin{cases}f\left(2\right)+3f\left(\frac{1}{2}\right)=4\\f\left(\frac{1}{2}\right)+3f\left(2\right)=\frac{1}{4}\end{cases}\Rightarrow f\left(2\right)=-\frac{13}{32}}\)
\(f\left(2\right)+3f\left(\frac{1}{2}\right)=4\) (1)
\(f\left(\frac{1}{2}\right)+3f\left(2\right)=\frac{1}{4}\)\(\Leftrightarrow\)\(3f\left(\frac{1}{2}\right)+9f\left(2\right)=\frac{3}{4}\) (2)
(1) - (2) \(\Leftrightarrow\)\(f\left(2\right)+3f\left(\frac{1}{2}\right)-3f\left(\frac{1}{2}\right)-9f\left(2\right)=4-\frac{3}{4}\)
\(\Leftrightarrow\)\(-8f\left(2\right)=\frac{13}{4}\)\(\Leftrightarrow\)\(f\left(2\right)=\frac{-13}{32}\)
Ta có : f(2) = f(2) + 3f(1/2) = 22 = 4. (1)
f(1/2) = f(1/2) + 3f( 1 / 1/2) =(1/2)2
= f(1/2) + 3f(2) =1/4 . (2)
= 3f(2) + f(1/2) = 1/4
= 9f(2) . 3f(1/2) = 1/4 . (2)
Lấy (2) trừ đi (1) ta có :
8f(2) = 3/4 -4 = -13/4
=> f(2) = -13/4 : 8 =-13/4 . 1/8 = -13/32
Vậy f(2) = -13/32
b/ Theo đề bài thì ta có:
\(\left\{{}\begin{matrix}f\left(1\right)=f\left(-1\right)\\f\left(2\right)=f\left(-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a_4+a_3+a_2+a_1+a_0=a_4-a_3+a_2-a_1+a_0\\16a_4+8a_3+4a_2+2a_1+a_0=16a_4-8a_3+4a_2-2a_1+a_0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a_3+a_1=0\\4a_3+a_1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a_3=0\\a_1=0\end{matrix}\right.\)
Ta có: \(f\left(x\right)-f\left(-x\right)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0-\left(a_4x^4-a_3x^3+a_2x^2-a_1x+a_0\right)\)
\(=2a_3x^3+2a_1x=0\)
Vậy \(f\left(x\right)=f\left(-x\right)\)với mọi x
a/ Áp dụng tính chất dãy tỷ số bằng nhau ta có:
\(\dfrac{a}{2015}=\dfrac{b}{2016}=\dfrac{c}{2017}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}=\dfrac{c-a}{2}\)
\(\Rightarrow c-a=-2\left(a-b\right)=-2\left(b-c\right)\)
Thế vào B ta được
\(B=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(=4\left(a-b\right)\left(b-c\right)-\left[-2\left(a-b\right).\left(-2\right).\left(b-c\right)\right]\)
\(=4\left(a-b\right)\left(b-c\right)-4\left(a-b\right)\left(b-c\right)=0\)