LH

Bài 8: Phân tích đa thức thành nhân tử.

a, x^4 - y^4

b, x^2 - 3y^2

c, (3x - 2y)^2 - (2x - 3y)^2

d, 9(x -y)^2 - 4(x + y)^2

e, (4x^2 - 4x + 1) - (x+1)^2

f, x^3 + 27

g, 27x^3 - 0,001

h, 125x^3 - 1

H9
22 tháng 8 2023 lúc 10:10

a) \(x^4-y^4\)

\(=\left(x^2\right)^2-\left(y^2\right)^2\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)

\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)

\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)

\(=\left(5x-5y\right)\left(x+y\right)\)

\(=5\left(x-y\right)\left(x+y\right)\)

d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)

\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)

\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)

\(=\left(5x-y\right)\left(x-5y\right)\)

e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)

\(=\left(2x-1\right)^2-\left(x+1\right)\)

\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)

\(=3x\left(x-2\right)\)

f) \(x^3+27\)

\(=x^3+3^3\)

\(=\left(x+3\right)\left(x^2-3x+9\right)\)

g) \(27x^3-0,001\)

\(=\left(3x\right)^3-\left(0,1\right)^3\)

\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)

h) \(125x^3-1\)

\(=\left(5x\right)^3-1^3\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)

Bình luận (0)