)1) x²-3√2x+4=0 2)x²+2(1+√8)x+8√2=0 3) x²-3x-5-√7=0 - вопрос №231186513240221882033570 от katyaydalova15 13.08.2021 18:28

Question

Алгебра: )1) x²-3√2x+4=0 2)x²+2(1+√8)x+8√2=0 3) x²-3x-5-√7=0

katyaydalova15 · Answer

1)x²-3√2x+4=0D=(-3√2)²-4*1*4=18-16=2=√2x1=(-(-3√2)+√2)/2*1=(3√2+√2)/2=4√2/2x1=2√2x2=(-(3√2)-√2)/2=(3√2-√2)/2=2√2/2x2=√22)D=(2+√32)²-4*1*8√2=4+4√32+32-32√2=36+16√2-32√2=(36-16√2)=√(36-22,627)==√13,373=3,657x1=(-(2+2√8)+3,657)/2*1=(-2-2√8+3,657)/2=-1,9999≈-2x2=(-2-2√8-3,657)/2=(-11,314)/2≈-5,6573) D=(-3)²-4*1*(-(5-√7)=9+20-4√7=29-4√7=29-10,583=18,417=√18,417≈4,292x1=(-(-3)+4,292)/2*1=7,292/2x1≈3,646x2=(-(-3)-4,292)/2=-1,292/2x2≈0,646...