Gui ban Duyen ve bai toan tim m de do thi hs cat Ox tai 3 diem phan biet co hoanh do lap thanh cap so cong?

De do thi hs cat Ox tai 3 diem phan biet M(x1;0); N(x2;0); P(x3;0) sao chho x1; x2; x3 theo thu tu do lap thanh CSC (khi do N la trung diem MP <=> x2 = (x1 + x3)/2 ) thi phai co dong thoi 2 dieu kien sau:

Dieu kien 1 : Hs co 2 cuc tri <=> pt y' = 0 co 2 nghiem phan biet

y' = 3x^2 - 6mx + 9 = 3(x^2 - 2mx + 3) = 0 (1)

Delta' = m^2 - 3 > 0 <=> m < - can3; m > can3 (*)

Goi diem cuc dai la A(a; y(a)); B(b; y(b)) . Theo Vi et ta co:

{ a + b = 2m (2)

{ ab = 3 (3)

Dieu kien 2 : 2 diem cuc tri nam ve 2 phia truc Ox <=> y(a); y(b) trai dau nhau va diem uon ( tuc tam doi xung) cua do thi chinh la N, khi do A cung doi xung voi B qua N nen : y(a) = - y(b)

Voi : y(a) = a^3 - 3ma^2 + 9a - 7 = (a^3 - 2ma^2 + 3a) - ma^2 + 6a - 7 = a(a^2 - 2ma + 3) - m(2ma - 3) + 6a - 7 = - 2(m^2 - 3)a + 3m - 7

(Vi a la nghiem cua pt (1) nen a^2 - 2ma + 3 = 0 => a^2 = 2m - 3)

Tuong tu : y(b) = - 2(m^2 - 3)b + 3m - 7

=> - 2(m^2 - 3)a + 3m - 7 = 2(m^2 - 3)b - 3m + 7

<=> 2(m^2 - 3)(a + b) - 6m + 14 = 0

<=> 2m^3 - 9m + 7 = 0 ( theo (2) co : a + b = 2m)

<=> (m - 1)(2m^2 + 2m - 7) = 0

Giai ra m = - (1 + can15)/2 ( loai nghiem m = 1; m = (can15 - 1)/2 vi khong thoa (*))