Question

giúp mình bài 3 ạ :3333

Answer 1

Answer 2

Answer 3

Bài 3,

x³ - 8y³ = (x - 2y)(x² + 2xy + 4y²)

8x³ - \(\dfrac{1}{125}\)y³ = \(\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

6x² + 7x - 5 = (2x - 1)(3x + 5)

x² + 10x + 24 = (x + 6)(x + 4)

\(\dfrac{1}{2}x^3+4\) = \(\dfrac{1}{2}\left(x^3+8\right)=\dfrac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)

x⁴ + 10x³ + 25x² = x²(x² + 10x + 25) = x²(x + 5)² = (x² + 5x)²

49x³ - 14x²y + xy² = x(49x² - 14xy + y²) = x. (7x - y)²

 

Answer 4

Bài 3: 

a: \(x^3-8y^3=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)

b: \(8x^3-\dfrac{1}{125}y^3=\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{5}{2}xy+\dfrac{1}{25}y^2\right)\)

c: \(6x^2+7x-5=6x^2+10x-3x-5\)

\(=\left(3x+5\right)\left(2x-1\right)\)

d: \(x^2+10x+24=\left(x+4\right)\left(x+6\right)\)

e: \(\dfrac{1}{2}x^3+4=\dfrac{1}{2}\left(x^3+8\right)=\dfrac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)

f: \(x^4+10x^3+25x^2=x^2\left(x^2+10x+25\right)=x^2\left(x+5\right)^2\)

g: \(49x^3-14x^2y+xy^2=xy\left(49x^2-14xy+y^2\right)=xy\left(7x-y\right)^2\)