Question

tìm x

(3x + 2) x (3x- 2)- (3x- 1)^2= 5

Answer 1

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

\(\Leftrightarrow\left(9x^2-2^2\right)-\left(9x^2-6x+1\right)=5\)

\(\Leftrightarrow9x^2-4-9x^2+6x-1-5=0\)
\(\Leftrightarrow6x=10\)

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

Vậy \(S=\left\{\dfrac{5}{3}\right\}\)

Answer 2

(3x + 2) . (3x- 2)- (3x- 1)^2= 5

<=> (3x + 2) . (3x- 2)- [ ( 3x^2 )  - 2 . 3x .1 + 1^2  ] = 5

<=>   9x^2 - 6x + 6x - 4 - ( 9x^2 - 6x + 1 ) = 5

<=>    9x^2 - 6x + 6x - 4 - 9x^2 + 6x - 1 = 5

<=>  6x - 5 = 5

<=> 6x = 5 + 5

<=> 6x = 10

<=> x = 10/6

<=> x = 5/3