(3x+2)(x-4)=0 5x^2+8x-4=0 25x^2-4=0 (x-2)^2-1=0 - вопрос №68661153240528402524022106218 от artem705 18.10.2021 09:51

Question

Алгебра: (3x+2)(x-4)=0 5x^2+8x-4=0 25x^2-4=0 (x-2)^2-1=0 6x^2=18x , с решением, )

artem705 · Answer

3x^2-10x-8=0x1;2=[10+-sqrt(100+96)]/6x1=23/6; x2=-1/25x^2+8x-4=0x1;2=[-8+-sqrt(64+80)]/10x1=2/5=0,4; x2=8/5=1,625x^2-4=0x^2=0,16x=sqrt(0,16)x=+-0,4(x-2)^2-1=0x^2-4x+4-1=0x^2-4x+3=0x1;2=[4+-sqrt(16-12)]/2x1=3; x2=16x^2=18x6x^2-18x=06x(x-3)=0x=0. x-3=0 x=3...