Lời giải:

Ta có:

\(f(x)=-5x\Rightarrow \left\{\begin{matrix} f(x_1)=-5x_1\\ f(x_2)=-5x_2\end{matrix}\right.\)

\(\Rightarrow f(x_1)-f(x_2)=-5x_1-(-5)x_2=-5(x_1-x_2)=5(x_2-x_1)\)

Do \(x_2> x_1\Rightarrow 5(x_2-x_1)>0\Leftrightarrow f(x_1)-f(x_2)>0 \)

\(\Leftrightarrow f(x_1)> f(x_2)\) (đpcm)

b)

\(\left\{\begin{matrix} f(x_1)=-5x_1\\ f(x_2)=-5x_2\rightarrow 4f(x_2)=-20x_2\end{matrix}\right.\)

\(\Rightarrow f(x_1)+4f(x_2)=-5x_1+(-20)x_2=-5x_1-20x_2\) (1)

Lại có:

\(f(x)=-5x\rightarrow f(x_1+4x_2)=-5(x_1+4x_2)=-5x_1-20x_2\) (2)

Từ (1),(2) suy ra \(f(x_1+4x_2)=f(x_1)+4f(x_2)\)

c)

\(f(x)=-5x\Rightarrow -f(x)=-(-5x)=5x\)

\(f(x)=-5x\Rightarrow f(-x)=-5(-x)=5x\)

Do đó: \(-f(x)=f(-x)\)

Câu 1: 

a) 

b) Ta có: f(x)=8

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

Vậy: Để f(x)=8 thì \(x\in\left\{2;-2\right\}\)

Ta có: \(f\left(x\right)=6-4\sqrt{2}\)

\(\Leftrightarrow2x^2=6-4\sqrt{2}\)

\(\Leftrightarrow x^2=3-2\sqrt{2}\)

\(\Leftrightarrow x=\sqrt{3-2\sqrt{2}}\)

hay \(x=\sqrt{2}-1\)

Vậy: Để \(f\left(x\right)=6-4\sqrt{2}\) thì \(x=\sqrt{2}-1\)

2: ĐKXĐ: x<>1

\(f'\left(x\right)=\dfrac{\left(x^2-3x+3\right)'\left(x-1\right)-\left(x^2-3x+3\right)\left(x-1\right)'}{\left(x-1\right)^2}\)

\(=\dfrac{\left(2x-3\right)\left(x-1\right)-\left(x^2-3x+3\right)}{\left(x-1\right)^2}\)

\(=\dfrac{2x^2-5x+3-x^2+3x-3}{\left(x-1\right)^2}=\dfrac{x^2-2x}{\left(x-1\right)^2}\)

f'(x)=0

=>x^2-2x=0

=>x(x-2)=0

=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

1:

\(f\left(x\right)=\dfrac{1}{3}x^3-2\sqrt{2}\cdot x^2+8x-1\)

=>\(f'\left(x\right)=\dfrac{1}{3}\cdot3x^2-2\sqrt{2}\cdot2x+8=x^2-4\sqrt{2}\cdot x+8=\left(x-2\sqrt{2}\right)^2\)

f'(x)=0

=>\(\left(x-2\sqrt{2}\right)^2=0\)

=>\(x-2\sqrt{2}=0\)

=>\(x=2\sqrt{2}\)

a: f(-2)-g(1/2)

\(=5\left(-2\right)-3+4\cdot\dfrac{1}{2}-1\)

\(=-10-4+2=-10-2=-12\)

b: \(2\cdot f^2\left(-3\right)-3\cdot g^2\left(-2\right)\)

\(=2\cdot\left[5\cdot\left(-3\right)-3\right]^2-3\cdot\left[\left(-4\right)\left(-2\right)+1\right]^2\)

\(=2\cdot\left(-18\right)^2-3\cdot9^2\)

\(=648-3\cdot81=405\)

c: Ở hai hàm số trên, nếu lấy biến x cùng một giá trị thì f(x) sẽ nhỏ hơn g(x) 3 đơn vị

a)  Cho hàm số y = f(x) = -2x + 3.

Ta có: f(-2)= -2.(-2)+3

                 = 4+3=7

Ta có: f(0)= -2.0+3

                = 0+3=3

Ta có: f(

Lời giải:

a.

$f(-2)=(-2)(-2)+3=7$

$f(0)=(-2).0+3=3$

$f(\frac{-1}{2})=(-2).\frac{-1}{2}+3=4$

b.

$f(x)=-2x+3=5$

$\Rightarrow -2x=2$

$\Rightarrow x=-1$

$f(x)=-2x+3=1$

$\Rightarrow -2x=1-3=-2$

$\Rightarrow x=1$