Question

[Trần Thị Thu Hường]

( y^4 - 2y^2- 8 ) : ( y - 2 ) = 0

Answer 1

[Nguyễn Huy Tú]

\(\frac{y^4-2y^2-8}{y-2}=\frac{\left(y^2-4\right)\left(y^2+2\right)}{y-2}=\frac{\left(y-2\right)\left(y+2\right)\left(y^2+2\right)}{y-2}\)

\(=\left(y+2\right)\left(y^2+2\right)=0\)

TH1 : \(y=-2\)

TH2 : vô nghiệm 

Vậy y = -2 

Answer 2

[l҉o҉n҉g҉ d҉z҉]

( y⁴ - 2y² - 8 ) : ( y - 2 ) = 0

<=> [ ( y⁴ - 2y² + 1 ) - 9 ] : ( y - 2 ) = 0

<=> [ ( y² - 1 )² - 3² ] : ( y - 2 ) = 0

<=> ( y² - 1 - 3 )( y² - 1 + 3 ) : ( y - 2 ) = 0

<=> ( y² - 4 )( y² + 2 ) : ( y - 2 ) = 0

<=> ( y - 2 )( y + 2 )( y² + 2 ) : ( y - 2 ) = 0

<=> ( y + 2 )( y² + 2 ) = 0

<=> y + 2 = 0 hoặc y² + 2 = 0

<=> y = -2 ( y² + 2 ≥ 2 > 0 ∀ y )