(x - 2/7)(x + 1/4) > 0

Xét 2 trường hợp:

                                                 Vậy x > 2/7 hoặc x < -1/4

\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\)  - 3) = \(x\)

\(\dfrac{1}{2}\)  \(\times\) \(\dfrac{3x-2}{3}\) -   \(\dfrac{2x-3}{3}\) = \(x\)

\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)

3\(x-2-4x\) + 6 = 6\(x\) 

-\(x\) + 4 - 6\(x\) = 0

7\(x\)  = 4

    \(x\) =  \(\dfrac{4}{7}\) 

2 (x-1) - 5 (x+2) = -10

2x-2 - 5x+10 = -10

2x-5x-2+10=-10

2x-5x=-10-10+2

-3x=-18

x=6

kho lam cung de

( x - 6 )( x + 1 ) = 0

<=> x - 6 = 0 hoặc x + 1 = 0

<=> x = 6 hoặc x = -1

     \(x^3-7x^2-13x+91=0\)

\(\Rightarrow x^2\left(x-7\right)-13\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(x^2-13\right)=0\)

\(\Rightarrow\left(x-7\right)\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)

Tìm được \(x\in\left\{7;\sqrt{13};-\sqrt{13}\right\}\)

x=2/45+1/9
x=7/45

Mình tưởng -\(\frac{2}{45}\)