Question

phân tích các đa thức sau thành nhân tử :

\(20x^7+7x-6\)

\(x^4-5x^2y^2+4y^4\)

\(x^8+y^4+1\)

Answer 1

Ta có: \(x^4-5x^2y^2+4y^4=\left(x^4-4x^2y^2+4y^4\right)-x^2y^2\)

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

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

Answer 2

1,

20x⁷ + 7x - 6

= 20x⁷ - 8x + 15x - 6

= (20x⁷ - 8x )+ (15x - 6)

= 4x( 5x - 2 ) + 3( 5x - 2)

= ( 5x - 2 )( 4x + 3 )

2/

x⁴ - 5x²y² + 4y⁴

= x⁴ - 4x²y² - x²y² + 4y²

= (x⁴ - 4x²y² + 4y²) - x²y²

= ( x² - 2y²)² - x²y²

= ( x² - 2y²)² - (xy)²

= ( x² - 2y² - xy)( x² - 2y² + xy)

3/

x⁸ + y⁴ + 1

= ( x⁸ - x² ) + ( x⁴ + 1 + x² )

= x²( x⁶ - 1) + ( x⁴ + 1 + x² )

= x²( x³ - 1 )( x³ + 1) + ( x⁴ + 1 + x² )

= x² . ( x - 1 ).( x² + x + 1 ). ( x - 1).(x² - x + 1) + ( x⁴ + 1 + x² )

= x² . \(\left[\left(x-1\right)\left(x+1\right)\right].\)\(\left[\left(x^2+x+1\right)\left(x^2-x+1\right)\right]\)+ ( x⁴ + 1 + x² )

= x²(x² - 1 )\(\left[\left(x^2+1\right)^2-x^2\right]\)+ ( x⁴ + 1 + x² )

= x²(x² - 1 )( x⁴ + 2x² + 1 - x² ) + ( x⁴ + 1 + x² )

= x²(x² - 1 )( x⁴ +x² + 1 ) + ( x⁴ + 1 + x² )

= ( x⁴ - x² )( x⁴ +x² + 1 ) + ( x⁴ + 1 + x² )

= ( x⁴ +x² + 1 )( x⁴ - x² + 1 )

Answer 3

Nếu đề là 20x⁷ + 7x - 6 thì là đề sai, phải ko?