x=-13/32

Theo c) \(f\left(\frac{5}{7}\right)=f\left(\frac{2}{7}+\frac{3}{7}\right)=f\left(\frac{2}{7}\right)+f\left(\frac{3}{7}\right)\)

      \(f\left(\frac{2}{7}\right)=f\left(\frac{1}{7}+\frac{1}{7}\right)=f\left(\frac{1}{7}\right)+f\left(\frac{1}{7}\right)=2.f\left(\frac{1}{7}\right)\) 

  \(f\left(\frac{3}{7}\right)=f\left(\frac{1}{7}+\frac{2}{7}\right)=f\left(\frac{1}{7}\right)+f\left(\frac{2}{7}\right)=f\left(\frac{1}{7}\right)+2f\left(\frac{1}{7}\right)=3.f\left(\frac{1}{7}\right)\)

       \(\implies\)\(f\left(\frac{5}{7}\right)=5.f\left(\frac{1}{7}\right)\)          (1)

 Theo b) \(f\left(\frac{1}{7}\right)=\frac{1}{7^2}.f\left(7\right)\)          (2)

Theo c) \(f\left(7\right)=f\left(3+4\right)=f\left(3\right)+f\left(4\right)\)

                                                 \(=2.f\left(3\right)+f\left(1\right)\) 

                                                 \(=6.f\left(1\right)+f\left(1\right)\) 

                                                 \(=7.f\left(1\right)\)

Theo a)\(f\left(1\right)=1\)\(\implies\)\(f\left(7\right)=7\)      (3)

    Từ (1);(2);(3)

       \(\implies\)       \(f\left(\frac{5}{7}\right)=\frac{5}{7}\)

︵✰He❤lloღ

thay x=2 và x=1/2 ta có 

\(\hept{\begin{cases}f\left(2\right)+3f\left(\frac{1}{2}\right)=4\\f\left(\frac{1}{2}\right)+3f\left(2\right)=\frac{1}{4}\end{cases}\Rightarrow f\left(2\right)=-\frac{13}{32}}\)

Từ giả thiết \(f\left(x_1+x_2\right)=f\left(x_1+x_2\right)\) ta có các biến đổi sau:

\(f\left(2020\right)=f\left(1024\right)+f\left(996\right)\)

\(=f\left(1024\right)+f\left(512\right)+f\left(484\right)\)

\(=f\left(1024\right)+f\left(512\right)+f\left(256\right)+f\left(228\right)\)

\(=f\left(1024\right)+f\left(512\right)+f\left(256\right)+f\left(128\right)+f\left(100\right)\)

\(=f\left(1024\right)+f\left(512\right)+f\left(256\right)+f\left(128\right)+f\left(64\right)\)

\(+f\left(36\right)\)

\(=f\left(1024\right)+f\left(512\right)+f\left(256\right)+f\left(128\right)+f\left(64\right)\)

\(+f\left(32\right)+f\left(4\right)\)

Dễ tính \(f\left(1024\right)=\)\(2.f\left(512\right)=4.f\left(256\right)=8.f\left(128\right)=16.f\left(64\right)\)

\(=32.f\left(32\right)=64.f\left(16\right)=128.f\left(8\right)=256.f\left(4\right)=512.f\left(2\right)\)

\(=1024.f\left(1\right)=1024\)

Tương tự ta có \(f\left(512\right)=512;f\left(256\right)=256;f\left(128\right)=128;f\left(64\right)=64;\)

\(f\left(32\right)=32;f\left(4\right)=4\)

\(\Rightarrow f\left(1024\right)+f\left(512\right)+f\left(256\right)+f\left(128\right)+f\left(64\right)\)

\(+f\left(32\right)+f\left(4\right)=2020\)

hay \(f\left(2020\right)=2020\)

Ta có: \(f\left(\frac{1}{x}\right)=\frac{1}{x^2}.f\left(x\right)\)

\(\Rightarrow f\left(\frac{1}{2020}\right)=\frac{1}{2020^2}.2020=\frac{1}{2020}\)

\(\Rightarrow f\left(\frac{3}{2020}\right)=f\left(\frac{2}{2020}\right)+f\left(\frac{1}{2020}\right)\)

\(=f\left(\frac{1}{2020}\right)+f\left(\frac{1}{2020}\right)+f\left(\frac{1}{2020}\right)\)

\(=\frac{1}{2020}.3=\frac{3}{2020}\)

Vậy \(f\left(\frac{3}{2020}\right)=\frac{3}{2020}\)

-13/32

Có :

\(3.f\left(2\right)+2.f\left(1-2\right)=2.2+9\)

\(\Rightarrow3.f\left(2\right)+2.f\left(-1\right)=13\)

\(3.f\left(-1\right)+2.f\left(2\right)=2.\left(-1\right)+9\)

\(\Rightarrow3.f\left(-1\right)+2.f\left(2\right)=7\)

\(\Rightarrow\left[3.f\left(2\right)+2.f\left(-1\right)\right]-\left[3.f\left(-1\right)+2.f\left(2\right)\right]=13-7\)

\(\Rightarrow f\left(2\right)-f\left(-1\right)=6\)

\(\Rightarrow f\left(-1\right)=f\left(2\right)-6\)

Thay \(f\left(-1\right)=f\left(2\right)-6\)vào \(3.f\left(2\right)+2.f\left(-1\right)=13\)có:

\(3.f\left(2\right)+2.\left[f\left(2\right)-6\right]=13\)

\(3.f\left(2\right)+2.f\left(2\right)-12=13\)

\(5.f\left(2\right)=25\)

\(f\left(2\right)=\frac{25}{5}=5\)

Vậy ...

thanks thùy dung nhiều

Cộng tác viên có khác mà

ra 5 nha bạn

Mình mới học lớp 6

Nên không biết nha

Chúc các bạn học giỏi

Ta có f(2)= 3f(2)+2f(-1)=2.2+9=13

f(-1)=3f(-1)+2f(2)=2.(-1)+9=7

=>f(-1)+f(2)=3f(2)+2f(-1)+3f(_1)+2f(2)=20

=:>5[f(2)+f(-1)]=20

=>f(2)+f(-1)=4

=>3f(2)+2f(_1)-3f(-1)-2f(2)=6

=>f(2)-f(-1)=6

=>f(2)+f(1)+f(2)+f(-1)=26

=>2f(2)=26

=>f(2)=13