Question

Phân tích đa thức thành nhân tử:

g) x⁴-16

i) -x²+10x-25

k) x³+3x²+3x+1-27z³

m) (x+y)² -25(x+y)+24

Answer 1

\(g,x^4-16=\left(x^2-4\right)\left(x^2+4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\\ i,-x^2+10x-25=-\left(x-5\right)^2\\ k,x^3+3x^2+3x+1-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\\ =\left(x-3z+1\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\\ m,\left(x+y\right)^2-25\left(x+y\right)+24=\left(x+y-5\right)^2-1=\left(x+y-4\right)\left(x+y-6\right)\)

Answer 2

g. x⁴ - 16

<=> x⁴ - 4²

<=> (x²)² - 4²

<=> (x² - 4)(x² + 4)

i. -x² + 10x - 25

<=> -(x² - 10x + 25)

<=> -(x² -10x + 5²)

<=> -(x - 5)²

Answer 3

sao ko ai like tui, buồn

Answer 4

g: \(x^4-16\)

\(=\left(x^2-4\right)\left(x^2+4\right)\)

\(=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)

h: \(-x^2+10x-25\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x-5\right)^2\)

k: \(x^3+3x^2+3x+1-27z^3\)

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

m: \(\left(x+y\right)^2-25\left(x+y\right)+24\)

\(=\left(x+y\right)^2-\left(x+y\right)-24\left(x+y\right)+24\)

\(=\left(x+y\right)\left(x+y-1\right)-24\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x+y-24\right)\)