138. Peminte hepaBeHCTBO ax2 + bxy + cy2 + dx + ky + p<0, ecdu: a) a = 2,6=c=0, d=k=-1, p=-4;
6) a= c = 1, b = 0, d=-4, k= 6, p = 9;
B) a = c = d=k=0, b = 2, p = -16.

Question

Алгебра: 138. Peminte hepaBeHCTBO ax2 + bxy + cy2 + dx + ky + p 0, ecdu: a) a = 2,6=c=0, d=k=-1, p=-4;6) a= c = 1, b = 0, d=-4, k= 6, p = 9;B) a = c = d=k=0, b = 2, p = -16.

aigamal · Answer

Объяснение:порове может быйт правилно...

138. Peminte hepaBeHCTBO ax2 + bxy + cy2 + dx + ky + p<0, ecdu: a) a = 2,6=c=0, d=k=-1, p=-4;6) a= c = 1, b = 0, d=-4, k= 6, p = 9;B) a = c = d=k=0, b = 2, p = -16.

138. Peminte hepaBeHCTBO ax2 + bxy + cy2 + dx + ky + p<0, ecdu: a) a = 2,6=c=0, d=k=-1, p=-4;
6) a= c = 1, b = 0, d=-4, k= 6, p = 9;
B) a = c = d=k=0, b = 2, p = -16.