Xây dựng ứng dụng giải và biện luận phương trình bậc 2 & giải toán tam giác
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phuong trinh da cho :
+ neu a = 0
hay : m+1 = 0 => m = -1
phuong trinh co 1 nghiem la: x = -1/2
+ neu a khac 0, hay :
+ Tinh delta, ta co delta = -4*m^2-12*m-4
+ neu delta = 0
m co nghiem: m1 = 1/2*5^(1/2)-3/2, m2=-3/2-
1/2*5^(1/2)
+ phuong trinh co nghiem kep tuong ung voi m la
x = -2/(5^(1/2)-1) voi m1 = 1/2*5^(1/2)-3/2
x = 2/(5^(1/2)+1) voi m2 = -3/2-1/2*5^(1/2)
+ neu delta > 0, hay 0 < -4*m^2-12*m-4
Trang9
=> m nam trong khoang sau:
phuong trinh co 2 nghiem phan biet x1, x2
x1 = -(1+(-m^2-3*m-1)^(1/2))/(m+1)
x2 = (-1+(-m^2-3*m-1)^(1/2))/(m+1)
+ neu delta < 0, hay -4*m^2-12*m < 4
=> m nam trong khoang sau thi phuong trinh vo
nghiem
>
phuong trinh da cho :
+ neu a = 0
hay : m+1 = 0 => m = -1
phuong trinh co 1 nghiem la: x = 0
+ neu a khac 0, hay :
+ Tinh delta, ta co delta = 4
+ delta = 4 > 0, phuong trinh co 2 nghiem phan
biet x1, x2
x1 = -2/(m+1)
x2 = 0
>
phuong trinh da cho :
+ neu a = 0
hay : m+1 = 0 => m = -1
phuong trinh vo nghiem
Trang10
+ neu a khac 0, hay :
+ Tinh delta, ta co delta = -36*m-36
+ neu delta = 0
m co nghiem: m = -1
+ phuong trinh co nghiem kep tuong ung voi m la
x = 0 voi m = -1
+ neu delta > 0, hay 0 < -36*m-36
=> m nam trong khoang sau:
phuong trinh co 2 nghiem phan biet x1, x2
x1 = 3/(-1-m)^(1/2)
x2 = -3/(-1-m)^(1/2)
+ neu delta < 0, hay -36*m < 36
=> m nam trong khoang sau thi phuong trinh vo
nghiem
>
phuong trinh da cho :
+ He so a khong co tham so m.
+ Tinh delta, ta co delta = m^2+4
+ neu delta = 0
truong hop nay vo nghiem m, nen phuong trinh da
cho khong co nghiem kep
+ neu delta > 0, hay 0 < m^2+4
=> Bat phuong trinh luon dung voi moi m
=> phuong trinh co 2 nghiem phan biet x1, x2
x1 = -1/2*m-1/2*(m^2+4)^(1/2)
x2 = -1/2*m+1/2*(m^2+4)^(1/2)
+ neu delta < 0, hay m^2 < -4
=> Bat phuong trinh vo nghiem, suy ra phuong
trinh da cho vo nghiem
>
Trang11
phuong trinh da cho :
+ He so a khong co tham so m.
+ Tinh delta, ta co delta = 25
+ delta = 25 > 0, phuong trinh co 2 nghiem phan
biet x1, x2
x1 = -2
x2 = 1/2
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