Question

1.Phân tích đa thức thành nhân tử bằng phương pháp tách,thêm hoặc bớt dạng ax²+bx+c :

a)x²-4x+2

b)x²-12x+11

c)3x²+6x-9

d)2x²-6x+2

Answer 1

a) x² - 4x + 2 = (x² - 4x + 4) - 2 = (x - 2)² - 2 = \(\left(x-2+\sqrt{2}\right)\left(x-2-\sqrt{2}\right)\)

b)  x² - 12x + 11 = x² - x - 11x + 11 = x(x - 1) - 11(x - 1) = (x - 1)(x - 11)

c) 3x² + 6x - 9 = 3x² - 3x + 9x - 9 = 3x(x - 1) + 9(x - 1) = (3x + 9)(x - 1) = 3(x + 3)(x - 1)

d) 2x² - 6x + 2 = 2(x² - 3x + 1) = 2(x² - 3x + 9/4 - 5/4) = 2[(x - 3/2)² - 5/4] = \(2\left(x-\frac{3}{2}+\sqrt{\frac{5}{4}}\right)\left(x-\frac{3}{2}-\sqrt{\frac{5}{4}}\right)\) 

Answer 2

1. 

a) \(x^2-4x+2=\left(x^2-4x+4\right)-2=\left(x-2\right)^2-2=\left(x-2-\sqrt{2}\right)\left(x-2+\sqrt{2}\right)\)

b) \(x^2-12x+11=\left(x^2-12x+36\right)-25=\left(x-6\right)^2-5^2=\left(x-6-5\right)\left(x-6+5\right)=\left(x-11\right)\left(x-1\right)\)

c) \(3x^2+6x-9=3\left(x^2+2x-3\right)=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)

d) \(2x^2-6x+2=2\left(x^2-3x+1\right)=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right]\)

\(=2\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)\)