Докажите тождество (1/(a + b + c)2 - 1/(a - b - c)2)(a2– - вопрос №12122210089168 от Mara1111176 12.01.2023 23:48

Question

Алгебра: Докажите тождество (1/(a + b + c)2 - 1/(a - b - c)2)(a2– (b+c)2)2/4a(b + c)=-1 ​

Mara1111176 · Answer

1tg(-a)*cosa+sina=-tga*cosa+sina=-sina*cosa/cosa +sina=-sina+sina=0 2 cos²a*tg²(-a)-1=cos²a*tg²a-1=cos²a*sin²a/cos²a-1=sin²a-1=-cos²a 3 ctg(-b)*sinb/cosb=-ctgb*sinb/cosb=-cosb*sinb/(sinb*cosb)=-1 4 (1-tg(-x))/(sinx+cos(-x))=(1+tgx)/(sinx+cosx)=(1+sinx/cosx)*1/(sinx+cosx)= =(cosx+sinx)/cosx*1/(sinx+cosx)=1/cosx 5 ctga*sin(-a)-cos(-a)=-ctga*sina-cosa=-cosa*sina/sina-cosa=-cosa-cosa= =-2cosa 6 tg(-u)ctgu+sin²u=-tgu*ctgu+sin²u=-1+sin²u=-cos²u 7 (1-sin²(-y))/(cosy=(1-sin²y)/cosy=cos²y/cosy=cosy 8 (tg(-x)+1)/(1-ctgx)=(-tgx+1)/(1-ctgx)=(-sinx/cosx+1):...

Докажите тождество (1/(a + b + c)2 - 1/(a - b - c)2)(a2– (b+c)2)2/4a(b + c)=-1 ​

Докажите тождество
(1/
(a + b + c)2 -
1/
(a - b - c)2)
(a2– (b+c)2)2/
4a(b + c)
=
-1