Question

Tìm x,biết

(6x+3)(3x-5)-(9x+5)(2x-3)=x+3

Answer 1

\(\left(6x+3\right)\left(3x-5\right)-\left(9x+5\right)\left(2x-3\right)=x+3\)

\(\Leftrightarrow18x^2-30x+9x-15-\left(18x^2-27x+10x-15\right)=x+3\)

\(\Leftrightarrow18x^2-30x+9x-15-18x^2+27x-10x+15-x-3=0\)

\(\Leftrightarrow-5x-3=0\)

\(\Leftrightarrow x=-\dfrac{3}{5}\)

Answer 2

(6x+3)(3x-5)-(9x+5)(2x-3)=x+3

( 18x² - 30x + 9x - 15 ) - ( 18x² - 27x+ 10x - 15 ) = x + 3

18x² - 30x + 9x - 15 - 18x² + 27x - 10x + 15 - x - 3 = 0

( 18x² - 18x² ) + ( - 30x + 9x + 27x - 10x - x ) + ( - 15 + 15 - 3 ) = 0

-5x - 3 = 0

-5x = 3

=> x = \(\dfrac{-3}{5}\)

Vậy x = \(\dfrac{-3}{5}\)