Đặt a = [tex]\frac{x - 2}{x + 1}[/tex] , b = [tex]\frac{x + 2}{x - 1}[/tex]
=> [tex]\frac{x^2 - 4}{x^2 - 1}[/tex] = ab
Ta có:
[tex]20a^2 - 5y^2 + 48xy[/tex] = 0
<=> [tex](20a^2 + 50ab) - (5y^2 + 2ab)[/tex] = 0
<=> 10a(2a + 5b) - b(2a + 5b) = 0
<=> (10a - b)(2a + 5b) = 0
<=> 10a - b = 0 hoặc 2a + 5b = 0
TH1: 10a - b = 0
<=> 10a = b
<=> [tex]\frac{10(x - 2)}{x + 1} = \frac{x + 2}{x - 1}[/tex]
<=> (x + 2)(x + 1) = 10(x - 2)(x - 1)
<=> [TEX]x^2 + 3x + 2[/TEX] = [TEX]10(x^2 - 3x + 2)[/TEX]
<=> [TEX]x^2 + 3x + 2[/TEX] = [TEX]10x^2 - 30x + 20[/TEX]
<=> [TEX]9x^2 - 33x + 18[/TEX] = 0
<=> [TEX]3x^2 - 11x + 6[/TEX] = 0
<=> x = 3 hoặc x = [tex]\frac{2}{3}[/tex]
TH2: 2a = 5b
<=> [tex]\frac{2(x - 2)}{x + 1} = \frac{5(x + 2)}{x - 1}[/tex]
<=> 2(x - 2)(x - 1) = 5(x + 2)(x + 1)
<=> [TEX]2(x^2 - 3x + 2)[/TEX] = [TEX]5(x^2 + 3x + 2[/TEX]
<=> [TEX]2x^2 - 6x + 4[/TEX] = [TEX]5x^2 + 15x + 10[/TEX]
<=> [TEX]3x^2 + 21x + 6[/TEX] = 0
<=> [TEX]x^2 + 7x + 2[/TEX] = 0
<=> x = [tex]\frac{-7 + \sqrt{41}}{2}[/tex] hoặc x = [tex]\frac{-7 - \sqrt{41}}{2}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
