a, \(Chof\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

- Lập bảng xét dấu :

Vậy \(\left\{{}\begin{matrix}f\left(x\right)>0\Leftrightarrow x\in\left(3;4\right)\\f\left(x\right)< 0\Leftrightarrow x\in\left(-\infty;3\right)\cup\left(4;+\infty\right)\\f\left(x\right)=0\Leftrightarrow x\in\left\{3;4\right\}\end{matrix}\right.\)

b, \(f\left(x\right)=\left(x-1\right)\left(x+6\right)\)

( Làm tương tự câu a )

1. Ta có : 3x+12=0 <=> x= -4

bảng xét dấu:

f(x) >0 ∀ x ∈ (-4;+∞)

f(x) <0 ∀ x∈ (-∞;-4)

2. Ta có : -5x+9=0 <=> x= \(\frac{9}{5}\)

Bảng xét dấu:

f(x) >0 ∀ x ∈ (-∞; 9/5)

f(x) <0 ∀ x ∈(9/5; +∞)

3. Ta có : -3x-9=0 <=> x= -3

f(x) >0 ∀ x∈ (-∞; -3)

f(x) <0 ∀x∈ ( -3; +∞ )

4. Ta có : x (2x+4)=0

+, x=0

+, 2x+4=0 <=> x= -2

f(x) >0 ∀ x ∈ (-∞; -2) \(\cup\) (0; +∞)

f(x) <0 ∀ x ∈ (-2;0)

5. Ta có: (x-2)(-x+4)=0

+, x-2=0 <=> x=2

+, -x+4=0 <=> x= 4

f(x) >0 ∀ x ∈ (2;4)

f (x) <0 ∀x∈ (-∞;2) \(\cup\)(4; +∞)

6. Ta có : (-4x+3)(x-6)=0

+, -4x+3=0 <=>x= \(\frac{3}{4}\)

+, x-6 =0 <=> x=6

f(x) >0 ∀ x∈ (3/4;6)

f(x) <0 ∀ x∈ (-∞; 3/4) \(\cup\)(6;+∞)

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

2: \(C=A:B\)

\(=\dfrac{x-1}{x^2}:\dfrac{x-1}{2x+1}=\dfrac{2x+1}{x^2}\)

\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)

=>C>=-1