4x⁴ - 21 x²y² + y⁴ 

= (4x⁴ + 4x²y² + y⁴) - 25x²y² 

=  [(2x²)² + 2x² . 2 . y² + (y²)²] - 25x²y²

= (2x² + y²) - 25x²y² 

= (2x² + y² - 5xy) (2x² + y² + 5xy)

bạn ơi cái chỗ 21x^2 ? y^2 là dấu cộng hay trừ vậy?

Nó là 1 á

4x⁴ - 21 x²y² + y⁴ 

= (4x⁴ + 4x²y² + y⁴) - 25x²y² 

=  [(2x²)² + 2x² . 2 . y² + (y²)²] - 25x²y²

= (2x² + y²) - 25x²y² 

= (2x² + y² - 5xy) (2x² + y² + 5xy)

4x² - 4y²

= (2x)² - (2y)²

= (2x - 2y)(2x + 2y)

(2x - 2y)(2x + 2y)

4x²-4y²=4.(x²-y²)=4.(x-y).(x+y)

\(=4x\left(x+2y\right)+3\left(x+2y\right)=\left(x+2y\right)\left(4x+3\right)\)

\(4x^2+8xy+3x+6y\)

\(=4\left(x+2y\right)+3\left(x+2y\right)\)

\(=7\left(x+2y\right)\)

=4x(x+2y)+3(x+2y)=(x+2y)(4x+3)

a) \(4x^4-21x^2y^2+y^4\)

Ấn nhầm :v

a) \(4x^4-21x^2y^2+y^4\)

\(=\left(2x^2\right)^2-2\cdot2x^2\cdot y^2+y^2-25x^2y^2\)

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

\(=\left(2x^2-5xy-y^2\right)\left(2x^2+5xy-y^2\right)\)

b) \(x^5-5x^3+4x\)

\(=x^5-4x^3-x^3+4x\)

\(=x^3\left(x^2-4\right)-x\left(x^2-4\right)\)

\(=\left(x^2-4\right)\left(x^3-x\right)\)

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

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

        \(4x^4-21x^2y^2+y^4\)

\(=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-25x^2y^2\)

\(=\left(2x^2+y^2\right)^2-\left(5xy\right)^2\)

\(=\left(2x^2-5xy+y^2\right)\left(2x^2+5xy+y^2\right)\)

       \(x^5-5x^3+4x\)

\(=x\left[x^4-5x^2+4\right]\)

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

\(=x\left[x^2\left(x-1\right)-4\left(x^2-1\right)\right]\)

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

Ta có: \(4x^2+12x+9\)

\(=4x^2+6x+6x+9\)

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

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

4x^4+4x^2y^2+y^4-25x^2y^2

=(2x^2+y^2)^2-(5xy)^2

=(2x^2+y^2-5xy)(2x^2+y^2+5xy)