Question

решить уравнение . (2х+1)^2- (4-2х)^2 =0

Answer 1

4x²+4x+1-(16-16x+4x²)=0
4x²+4x+1-16+16x-4x²=0
20x-15=0
20x=15
x=0,75

Answer 2

$(2x + 1) {}^{2} - (4 - 2x) {}^{2} = 0 \ 4x {}^{2} + 4x + 1 - (16 - 16x + 4x {}^{2} ) = 0 \ 4x {}^{2} + 4x + 1 - 16x - 4x {}^{2} = 0 \ 20x - 15 = 0 \ 20x = 15 div 20 \ x = frac{3}{4}$