Question

tìm x biết : \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)

 

Answer 1

\(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)

\(3x^2+2x+x^2+2x+1-4x^2+25=-12\)

\(4x+26=-12\)

\(4x=-38\)

\(x=\frac{-19}{2}\)

Answer 2

x(3x+2) + (x+1)² - (2x-5)(2x+5)= -12

(3x²+2x) + (x²+2x+1) - (4x² - 25) = -12

3x² + 2x + x² + 2x + 1 - 4x² +25 = -12

(3x² + x² - 4x²) + ( 2x+2x) + (1+25) = -12

0 + 4x + 26 = -12

4x = -12 - 26

4x = -38

x = -9.5