Question

Answer 1

12x -10 hay 2x-10 vậy

Answer 2

b) ⇔ \(9x^2=1\) ⇔ \(x^2 = \dfrac{1}{9} \)

⇔ \(x=±\dfrac{1}{3} \) 

Vậy...

c) \(⇔ 7x^2 - 35x=2x-10\) ⇔ \(7x^2-37x+10=0\)

⇔ \((7x^2-2x)-(35x-10)=0⇔x(7x-2)-5(7x-2)=0 \)

⇔ \((7x-2)(x-5)=0 \) ⇒  \(\left[\begin{array}{} 7x-2=0\\ x-5=0 \end{array} \right.\) ⇔ \(\left[\begin{array}{} x=\dfrac{2}{7} \\ x=5 \end{array} \right.\)

Vậy ...