[a,b=arctg],[b,c,bdot x+cy],[axby,bx+ay,0]|=(b^(2)-ac)(ax^(2)+2bxy+c)

Prove that: |[a, b, ax+by],[ b, c, bx+cy], [ax+by, bx+cy,0]|=(b^2-a c)(a x^2+2b x y+c y^2)

Prove that |(a,b,ax+by),(b,c,bx+cy),(ax+by, bx + cy, 0)| = (b^(2)-ac)(ax^(2) + 2bxy + cy^(2)) .

det[[ Prove that: ,b,ax+bya,c,bx+cyax+by,bx+cy,0]]=(b^(2)-ac)(ax^(2)+2bxy+cy^(2))

|{:(a,b,ax+by),(b,c,bx+cy),(ax+by,bx+cy,0):}|=(b^2-ac)(ax^2+2bxy+cy^2)

[[a, b, ax + byb, c, bx + cyax + by, bx + cy, 0]] = (b ^ (2) -ac) (ax ^ (2) + 2bxy + cy ^ (2))

Prove that: |(a,b, ax+by),(b,c,bx+cy), (ax+by, bx+cy,0)|=(b^2-a c)(a x^2+2b x y+c y^2) .

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0