Question

(3x-1)²-5(x+3)(x-3)=(2x+1)²+35

Answer 1

$\displaystyle (3x-1)^2-5(x+3)(x-3)=(2x+1)^2+35 \\ (3x)^2-2 \cdot 3x \cdot 1+1^2-5(x^2-9)=(2x)^2+2 \cdot 2x \cdot 1+1^2+35 \\ 9x^2-6x+1-5x^2+45=4x^2+4x+1+35 \\ 9x^2-5x^2-4x^2-6x-4x=1+35-1-45 \\ -10x=-10 \Rightarrow x= \frac{-10}{-10} \Rightarrow x=1$

Answer 2

(3x-1)²-5(x+3)(x-3)=(2x+1)²+35
[(3x)²-2·3x·1+1²]-5(x²-9)=[(2x)²+2·2x·1+1²)]+35
(9x²-6x+1)-5(x²-9)=(4x²+4x + 1 ) +35
9x²-6x+1-5x²+45=4x²+4x +1+35
9x²-6x+46-5x²=4x²+4x+36
9x²-6x-5x²-4x²-4x=+36-46    ....... ...... 9x²-4x²-5x²=0(nu se mai scrie de acum incolo in ecuatie.) ; -6x-4x= -10 x 
⇒    -10x= -10 /·(-1)   ...... se inmulteste cu -1 pentru a face ecuatia fara - . 
10x=10
x=10/10
x=1
Sper ca te-am ajutat!