Bài 39. Giải phương trình bằng phương pháp đưa về phương trình tích.
a) ((3x^2 – m 7x m - m 10)<2x^2 + m left( 1 m – m sqrt 5 ight)x m + m sqrt 5 m - m 3> m = m 0);
b) (x^3 + m 3x^2- m 2x m - m 6 m = m 0);
c) ((x^2 – m 1)left( 0,6x m + m 1 ight) m = m 0,6x^2 + m x);
d) ((x^2 + m 2x m - m 5)^2 = m ( m x^2- m x m + m 5)^2).
a) ((3x^2 – m 7x m - m 10)<2x^2 + m left( 1 m – m sqrt 5 ight)x m + m sqrt 5 m - m 3> m = m 0)
(Leftrightarrow)(left< matrix (3x^2 – m 7x m - m 10) m = m 0(1) hfill cr 2x^2 + m left( 1 m – m sqrt 5 ight)x m + sqrt 5 - m 3 m = m 0(2) hfill cr ight.)
Giải (1): phương trình (a – b + c = 3 + 7 – 10 = 0)
nên (x_1 = – 1,x_2 = – – 10 over 3 = 10 over 3)
Giải (2): phương trình có (a + b + c = 2 + (1 – sqrt5) + sqrt5 – 3 = 0)
nên (x_3 = 1,x_4 = sqrt 5 – 3 over 2)
b) (x^3 + m 3x^2- m 2x m - m 6 m = m 0) (Leftrightarrow x^2left( x m + m 3 ight) m - m 2left( x m + m 3 ight) m = m 0 )
(Leftrightarrow left( x m + m 3 ight)(x^2 – m 2) m = m 0)
(Leftrightarrow)(left< matrix x + 3 = 0 hfill cr x^2 – m 2 m = m 0 hfill cr ight.)
Giải ra (x_1 = m – 3, m x_2 = m – sqrt 2 , m x_3 = sqrt 2 )
c) ((x^2 – m 1)left( 0,6x m + m 1 ight) m = m 0,6x^2 + m x) ( Leftrightarrow m left( 0,6x m + m 1 ight)left( x^2- m x m - m 1 ight) m = m 0)Quảng cáo
(Leftrightarrow left< matrix 0,6x + 1 = 0(1) hfill cr x^2- m x m - m 1 m = m 0(2) hfill cr ight.)
(1) ⇔ (0,6x + 1 = 0 )
( Leftrightarrow x_1 = – 1 over 0,6 = – 5 over 3)
(2):(Delta = ( – 1)^2 – 4.1.( – 1) = 1 + 4 = 5,sqrt Delta = sqrt 5,)
(x_2 = m 1 – sqrt 5 over 2,x_3 = 1 + sqrt 5 over 2)
Vậy phương trình có cha nghiệm:
(x_1 = – 5 over 3,x_2 = 1 – sqrt 5 over 2,x_3 = 1 + sqrt 5 over 2),
d) ((x^2 + m 2x m - m 5)^2 = m ( m x^2- m x m + m 5)^2)( Leftrightarrow m (x^2 + m 2x m - m 5)^2 – m ( m x^2- m x m + m 5)^2 = m 0)
(Leftrightarrow (x^2 + m 2x m - m 5 m + m x^2- m x m + m 5).)
(( m x^2 + m 2x m - m 5 m – m x^2 + m x m – m 5) m = m 0)
( Leftrightarrow m (2x^2 + m x)left( 3x m - m 10 ight) m = m 0)