Cho A(x) = $x^{3}$ – 4x B(x) = $x^{3}$ – $x^{2}$ – 12x – 15 Tìm x sao cho A(x) = B(x) 11/08/2021 Bởi Piper Cho A(x) = $x^{3}$ – 4x B(x) = $x^{3}$ – $x^{2}$ – 12x – 15 Tìm x sao cho A(x) = B(x)
A(x) = x³ – 4x B(x) = x³ – x² – 12x – 15 A(x) = B(x) ⇔ x³ – 4x = x³ – x² – 12x – 15 ⇔ x² + 12x – 4x + 15 = 0 ⇔ x² + 8x + 15 = 0 ⇔ x² + 5x + 3x + 15 = 0 ⇔ (x² + 5x ) + (3x + 15 )= 0 ⇔ x (x + 5 ) + 3 ( x + 5 ) = 0 ⇔ ( x + 3 ) ( x + 5 ) = 0 ⇔ \(\left[ \begin{array}{l}x=-5\\x=-3\end{array} \right.\) Vậy để A(x) = B(x) thì x ∈ { – 5 ; – 3 } Bình luận
`A(x) = B(x)`hay `x^3 – 4x = x^3 – x^2 – 12x – 15``<=> x^3 – 4x – x^3 + x^2 + 12x + 15= 0``<=> x^2 + 8x + 15 = 0``<=> x^2 + 3x + 5x + 15 = 0``<=> x( x + 3 ) + 5( x + 3 ) = 0``<=> ( x + 3 )( x + 5 ) = 0``<=>` \(\left[ \begin{array}{l}x=-3\\x=-5\end{array} \right.\) Vậy để `A( x ) = B(x)` thì `x = -3` hoặc `x = -5` Bình luận
A(x) = x³ – 4x
B(x) = x³ – x² – 12x – 15
A(x) = B(x)
⇔ x³ – 4x = x³ – x² – 12x – 15
⇔ x² + 12x – 4x + 15 = 0
⇔ x² + 8x + 15 = 0
⇔ x² + 5x + 3x + 15 = 0
⇔ (x² + 5x ) + (3x + 15 )= 0
⇔ x (x + 5 ) + 3 ( x + 5 ) = 0
⇔ ( x + 3 ) ( x + 5 ) = 0
⇔ \(\left[ \begin{array}{l}x=-5\\x=-3\end{array} \right.\)
Vậy để A(x) = B(x) thì x ∈ { – 5 ; – 3 }
`A(x) = B(x)`
hay `x^3 – 4x = x^3 – x^2 – 12x – 15`
`<=> x^3 – 4x – x^3 + x^2 + 12x + 15= 0`
`<=> x^2 + 8x + 15 = 0`
`<=> x^2 + 3x + 5x + 15 = 0`
`<=> x( x + 3 ) + 5( x + 3 ) = 0`
`<=> ( x + 3 )( x + 5 ) = 0`
`<=>` \(\left[ \begin{array}{l}x=-3\\x=-5\end{array} \right.\)
Vậy để `A( x ) = B(x)` thì `x = -3` hoặc `x = -5`