vậy f(1/2)+3.f(2)=1/4 hay 3f(1/2)+9.f(2)=3/4

và f(2)+3.f(1/2)=4 

trừ vế theo vế ta đc 

8.f(2)=-13/4

suy ra f(2)=-13/32

mình ko biết xin lỗi bạn nha!

mình ko biết xin lỗi bạn nha!

mình ko biết xin lỗi bạn nha!

mình ko biết xin lỗi bạn nha!

mình ko biết xin lỗi bạn nha!

\(f\left(\frac{1}{3}\right)+2f\left(\frac{1}{\frac{1}{3}}\right)=\left(\frac{1}{3}\right)^2\Rightarrow f\left(\frac{1}{3}\right)+2f\left(3\right)=\frac{1}{9}\)(1)

\(f\left(3\right)+2f\left(\frac{1}{3}\right)=3^2\Rightarrow2f\left(3\right)+4f\left(\frac{1}{3}\right)=18\)(2)

Từ (1) và (2) \(\Rightarrow2f\left(3\right)+4f\left(\frac{1}{3}\right)-f\left(\frac{1}{3}\right)-2f\left(3\right)=18-\frac{1}{9}\)

\(\Rightarrow3f\left(\frac{1}{3}\right)=\frac{161}{9}\Rightarrow f\left(\frac{1}{3}\right)=\frac{161}{27}\)

Thay 1/3 vào x là xong

NGU VL

Xét hàm số f(x) thỏa mãn f(x)+2f(1/x)=x^2. với mọi x thuộc R.
Đúng với x = 2 . => f(2) + 2f(1/2) = 2^2 = 4
=> f(2) + 2f(1/2) = 4 ( 1 )
Đúng với x = 1/2 => f(1/2) + 2f(2) = (1/2)^2 = 1/4.
=> 2f(2) + f (1/2) = 1/4.=> 4f(2) + 2f(1/2) = 2/4 ( 2 )
Lấy (2) trừ (1) ta đc :  3f(2) = 2/4 - 4 = -7/2
=> f(2) = -7/2: 3= -7/6

Với x=2 ta có \(f\left(2\right)-3f\left(\frac{1}{2}\right)=4\left(1\right)\)

Với x=1/2 ta có:\(f\left(\frac{1}{2}\right)-3f\left(2\right)=\frac{1}{4}\Rightarrow3f\left(\frac{1}{2}\right)-9f\left(2\right)=\frac{3}{4}\left(2\right)\)

Lấy (1) cộng (2) ta có

\(\Rightarrow f\left(2\right)-3f\left(\frac{1}{2}\right)+3f\left(\frac{1}{2}\right)-9f\left(2\right)=4+\frac{3}{4}\)

\(\Rightarrow-8f\left(2\right)=\frac{19}{4}\)

\(\Rightarrow f\left(2\right)=-\frac{19}{32}\)

\(f\left(x\right)=\frac{x^2+2x+1-x^2}{x^2\left(x+1\right)^2}=\frac{\left(x+1\right)^2-x^2}{x^2\left(x+1\right)^2}=\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}\)

\(\Rightarrow f\left(1\right)+f\left(2\right)+....+f\left(x\right)=1-\frac{1}{2^2}+\frac{1}{2^2}-....-\frac{1}{\left(x+1\right)^2}\)

\(\Rightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)

\(\Leftrightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-20+\left(x+1\right)=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)

Dat:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)

\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)

\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)

ta có: f(x) +3f(1/x) =x^2 với mọi x thuộc R

mà f(2)

=> f(2) +3f(1/2) = 2^2 =4 (1)

=> 3f(2) +f(1/2) =1/4 => 9f(2) +3f(1/2) = 3/4 (2)

=> (2) -(1) = 9f(2) +3f(1/2) - f(2) - 3f(1/2)

                 = 8f(2) = 3/4 -4

                            = -13/4

=> 8f(2) = -13/4

f(2)        = -13/4 :8

f(2)       = -13/32

p/s nha bn !!!!

\(f\left(x\right)+3f\left(\frac{1}{x}\right)=x^2\)

Tại x=2 \(\Rightarrow f\left(2\right)+3f\left(\frac{1}{2}\right)=2^2=4\left(1\right)\)

Tại x=\(\frac{1}{2}\)\(\Rightarrow f\left(\frac{1}{2}\right)+3f\left(2\right)=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

\(\Rightarrow3f\left(\frac{1}{2}\right)+9f\left(2\right)=\frac{3}{4}\)

\(\Rightarrow9f\left(2\right)+3f\left(\frac{1}{2}\right)=\frac{3}{4}\left(2\right)\)

Từ (1)(2) \(\Rightarrow\hept{\begin{cases}f\left(2\right)+3f\left(\frac{1}{2}\right)=4\\9f\left(2\right)+3f\left(\frac{1}{2}\right)=\frac{3}{4}\end{cases}\Rightarrow8f\left(2\right)=\frac{3}{4}-4=\frac{-13}{4}}\)

\(\Rightarrow f\left(2\right)=\frac{-13}{32}\)

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)=