Toán Lớp 8: tìm giá trị cảu x b ) 3 ( x + 4 ) – x bình phương – 4x = 0
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Toán Lớp 8: tìm giá trị cảu x b ) 3 ( x + 4 ) – x bình phương – 4x = 0
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Giải đáp:
3.(x + 4) – x^2 – 4x = 0
↔ 3.(x + 4) – x.(x +4) = 0
↔ (x + 4).(3 – x) = 0
↔ $\left[\begin{matrix} x+ 4 = 0\\ 3 – x = 0\end{matrix}\right.$
↔ $\left[\begin{matrix} x=-4\\ x = 3\end{matrix}\right.$
Vậy x ∈ {-4;3}
$#dariana$
3 ( x + 4 ) – x² – 4x = 0
⇔ 3 . x + 3 . 4 – x² – 4x = 0
⇔ 3x + 12 – x² – 4x = 0
⇔ -x² + ( 3x – 4x ) + 12 = 0
⇔ -x² – x + 12 = 0
⇔ x² + x – 12 = 0
⇔ x² + 4x – 3x – 12 = 0
⇔ ( x² + 4x ) – ( 3x + 12 ) = 0
⇔ x ( x + 4 ) – 3 ( x + 4 ) = 0
⇔ ( x + 4 ) ( x – 3 ) = 0
⇔ $\left[\begin{matrix} x + 4 = 0 \\ x – 3 = 0 \end{matrix}\right.$
⇔ $\left[\begin{matrix} x = -4 \\ x = 3 \end{matrix}\right.$
Vậy x = -4 hoặc x = 3