Question

phân tích đa thức thành nhân tử: x^3-19x-30. Tìm x,biết. (x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0

Answer 1

x³ - 19x - 30

= x³ - 9x - 10x - 30

= x(x² - 9) - 10.(x + 3)

= x(x² - 3²) - 10.(x + 3)

= x.(x+3).(x-3) - 10.(x+3)

= (x+3).[x.(x-3) - 10]

= (x+3).(x² - 3x - 10)

= (x+3).(x² + 2x - 5x - 10)

= (x+3).[x(x+2) - 5(x+2)]

= (x+3).(x+2).(x-5).

Answer 2

Bài 1:

Ta có:

\(x^3-19x-30\\ =x^3-5x^2+5x^2-25x+6x-30\\ =x^2\left(x-5\right)+5x\left(x-5\right)+6\left(x-5\right)\\ =\left(x-5\right)\left(x^2+5x+6\right)\\ =\left(x-5\right)\left(x+2\right)\left(x+3\right)\)

Bafi2:

\(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\\ \Leftrightarrow x-2=0\left(dox^2+4x+6>0\right)\\ \Leftrightarrow x=2\)

Vậy x=2

Answer 3

Ta có: (x-2)(x² + 2x + 7) + 2(x² -4) - 5(x-2) = 0

(x-2)(x² + 2x + 7) + 2(x² - 2²) - 5(x-2) = 0

(x-2)(x² + 2x + 7) + 2(x-2)(x+2) - 5(x-2) = 0

(x-2). [(x² + 2x + 7) + 2(x+2) - 5] = 0

(x-2).[x² + 2x + 7 + 2x + 4 - 5] = 0

(x-2).[x² + 4x + 6] = 0

=> x-2 = 0 (x² + 4x + 6 > 0)

+) x - 2 = 0

=> x = 2

Vậy x = 2.