Với x=0

\(\Rightarrow3.f\left(0\right)-f\left(1\right)=0+1=1\)

\(f\left(0\right)-f\left(1\right)=\frac{1}{3}\)(1)

Với x=1

\(\Rightarrow3.f\left(1\right)-f\left(0\right)=1+1=2\)

\(f\left(1\right)-f\left(0\right)=\frac{2}{3}\)(2)

Với x=-1

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

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

Kết hợp (1);(2);(3) tính nhé

Lời giải:

Ta có: \(f(x)=ax^2+bx+c\)

\(\Rightarrow \left\{\begin{matrix} f(x+3)=a(x+3)^2+b(x+3)+c\\ f(x+2)=a(x+2)^2+b(x+2)+c\\ f(x+1)=a(x+1)^2+b(x+1)+c\\ f(x)=ax^2+bx+c\end{matrix}\right.\)

\(\Rightarrow f(x+3)-3f(x+2)+3f(x+1)-f(x)\)

\(=[f(x+3)-f(x)]-3[f(x+2)-f(x+1)]\)

Có:

\(f(x+3)-f(x)=a(x+3)^2+b(x+3)+c-[ax^2+bx+c]\)

\(=a[(x+3)^2-x^2]+b(x+3-x)\)

\(=3a(2x+3)+3b(1)\)

Và: \(f(x+2)-f(x+1)=a[(x+2)^2-(x+1)^2]+b[(x+2)-(x+1)]\)

\(=a(2x+3)+b\)

\(\Rightarrow 3[f(x+2)-f(x+1)]=3a(2x+3)+3b(2)\)

Từ (1)(2) suy ra:

\(f(x+3)-3f(x+2)+3f(x+1)-f(x)=3a(2x+3)+3b-[3a(2x+3)+3b]=0\)

-13/32

1: \(2\cdot f\left(x\right)+3\cdot f\left(5-x\right)=1-2x\)

=>\(2f(5-x)+3f(5-(5-x))=1-2(5-x)\)

=>\(2f(5-x)+3f(x)=2x-9\quad\)

=>\(9\cdot f\left(x\right)+6\cdot f\left(5-x\right)=6x-27\) (1)

\(2\cdot f\left(x\right)+3\cdot f\left(5-x\right)=1-2x\)

=>\(4\cdot f\left(x\right)+6\cdot f\left(5-x\right)=2-4x\) (2)

Từ (1),(2) suy ra \(9\cdot f\left(x\right)+6\cdot f\left(5-x\right)-4\cdot f\left(x\right)-6\cdot f\left(5-x\right)\) =6x-27-2+4x

=>\(5\cdot f\left(x\right)=10x-29\)

=>f(x)=2x-5,8

2: Ta có: \(4f\left(\frac{1}{x}\right) - 3f(x) = 12x^2\)

=>\(4f(x)-3f\left(\frac{1}{x}\right)=12\cdot\left(\frac{1}{x}\right)^2=\frac{12}{x^2}\quad\)

=>\(16f(x)-12f\left(\frac{1}{x}\right)=\frac{48}{x^2}\quad\) (1)

Ta có: \(-3f(x)+4f\left(\frac{1}{x}\right)=12x^2\quad\)

=>\(-9f(x)+12f\left(\frac{1}{x}\right)=36x^2\quad(2)\)

Từ (1),(2) suy ra \(16f(x)-12f\left(\frac{1}{x}\right)-9f(x)+12f\left(\frac{1}{x}\right)\) =\(\frac{48}{x^2}+36x^2\)

=>\(7\cdot f\left(x\right)=\frac{48}{x^2}+36x^2\)

=>\(f\left(x\right)=\frac{48}{7x^2}+\frac{36}{7}x^2\)