⇔ [( x + 2 )( x+12 )][( x + 3 )(x + 8)] = 4x²
⇔ ( x\(^2\) + 2x + 12x + 24 ) ( x\(^2\) + 3x + 8x + 24 ) = 4x²
Đặt x\(^2\) + 24 là a tacó :
pt⇔( a + 14x )( a + 11x ) = 4x\(^2\)
⇔ a\(^2\) + 11ax + 14ax + 154x\(^2\) - 4x\(^2\) = 0
⇔ a\(^2\) + 25ax + 150x\(^2\) = 0
⇔ a\(^2\) + 15ax + 10ax + 150x\(^2\) = 0
⇔ a( a + 15x ) + 10x ( a + 15x ) = 0
⇔ ( a + 10x ) ( a + 15x ) = 0
Thay a bằng x\(^2\) + 24
pt⇔ ( x\(^2\) + 24 + 10x ) ( x\(^2\) + 24 + 15x ) = 0
⇔ ( x\(^2\) + 4x + 6x + 24 ) ( x\(^2\) + 15x + 24 ) = 0
⇔ [ x( x + 4 ) + 6 (x + 4 )] ( np in dam) = 0
⇔ [ ( x + 6 ) ( x + 4 ) ] ( cnt ) = 0
⇔ \(\left[{}\begin{matrix}x+6=0\\x+4=0\\x^2+15x+24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-4\\x\approx-1,82\\x\approx-13,18\end{matrix}\right.\)
