a) theo tính chất  ta có: f(0+0)= f(0)+f(0)

=> f(0)=f(0)+f(0)

=> f(0)-f(0)=f(0)+f(0)-f(0)

=> 0=f(0)

hay f(0)=0

b)  f(0)=f(-x+x)=f(-x)+f(x)

=>0=f(-x)+f(x)

=> f(-x)=0-f(x)=-f(x)

c) \(f\left(x_1-x_2\right)=f\left(x_1+\left(-x_2\right)\right)=f\left(x_1\right)+f\left(-x_2\right)=f\left(x_1\right)-f\left(x_2\right)\)

a, f(10x) = k.(10x) = 10.(kx) = 10.f(x)

b, f(x₁ + x₂) = k(x₁ + x₂) = kx₁ + kx₂ = f(x₁) + f(x₂)

c, f(x₁ - x₂) = k(x₁ - x₂) = kx₁ - kx₂ = f(x₁) - f(x₂)

Từ x₁ < x₂ và 3 > 0

suy ra : 3x₁< 3x₂ hay f(x₁) < f(x₂ ).

Vậy hàm số đã cho đồng biến trên R.

ta có:

\(f\left(x_1\right)=kx_1;f\left(x_2\right)=kx_2=>f\left(x_1-x_2\right)=k.\left(x_1-x_2\right)=kx_1-kx_2\)

vậy \(f\left(x_1-x_2\right)=f\left(x_1\right)-f\left(x_2\right)\)

tick mk nhé

Bài 1:

nếu x₁<x₂=>2018.x₁-3<2018.x₂

=>f(x₁)<f(x₂)

Bài 2:

nếu x dương=>100x²+2 dương

nếu x âm=>100x²+2 dương vì  x² luôn dương

=>f(x)=f(-x)

Bài 3:

nếu x₁<x₂=>-2019x₁+1<2019x₂+1

=>f(x₁)<f(x₂)

\(f\left(\frac{5}{7}\right)=f\left(\frac{1}{\frac{7}{5}}\right)=\frac{1}{\left(\frac{7}{5}\right)^2}.f\left(\frac{7}{5}\right)=\frac{25}{49}.f\left(1+\frac{2}{5}\right)=\frac{25}{49}.\left(f\left(1\right)+f\left(\frac{2}{5}\right)\right)\)

Ta có : \(f\left(\frac{2}{5}\right)=f\left(\frac{1}{5}+\frac{1}{5}\right)=f\left(\frac{1}{5}\right)+f\left(\frac{1}{5}\right)=2.f\left(\frac{1}{5}\right)=2.\frac{1}{5^2}.f\left(5\right)=\frac{2}{25}.f\left(1+1+1+1+1\right)\)

\(=\frac{2}{25}.\left(f\left(1\right)+f\left(1\right)+f\left(1\right)+f\left(1\right)+f\left(1\right)\right)=\frac{2}{25}.5=\frac{2}{5}\)

Vậy \(f\left(\frac{5}{7}\right)=\frac{49}{25}.\left(1+\frac{2}{5}\right)=\frac{25}{49}.\frac{7}{5}=\frac{5}{7}\)