Question

(x+5)^2+(2x-3)^2-5(x-2) (x+2)=0

đây là bài tìm x ạ

 

Answer 1

\(x^2 + 10x + 25 + 4x^2 - 12x + 9 - 5(x^2 - 4) = 0\\ ⇔ -2x + 54 = 0\\ ⇔ x = 27\)

Answer 2

Ta có: \(\left(x+5\right)^2+\left(2x-3\right)^2-5\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow x^2+10x+25+4x^2-12x+9-5\left(x^2-4\right)=0\)

\(\Leftrightarrow5x^2-2x+34-5x^2+20=0\)

\(\Leftrightarrow-2x+54=0\)

hay x=27

Vậy: S={27}