Đặt x2 + 10x + 24 = y
pt đã cho trở thành ( y + 4x ).y - 165x2 = 0
<=> y2 + 4xy - 165x2 = 0
<=> y2 - 11xy + 15xy - 165x2 = 0
<=> y( y - 11x ) + 15x( y - 11x ) = 0
<=> ( y - 11x )( y + 15x ) = 0
=> ( x2 + 10x + 24 - 11x )( x2 + 10x + 24 + 15x ) = 0
<=> ( x2 - x + 24 )( x2 + 25x + 24 ) = 0
<=> ( x2 - x + 24 )( x2 + 24x + x + 24 ) = 0
<=> ( x2 - x + 24 )[ x( x + 24 ) + ( x + 24 ) ] = 0
<=> ( x2 - x + 24 )( x + 24 )( x + 1 ) = 0
Vì x2 - x + 24 > 0 ∀ x
nên pt <=> ( x + 24 )( x + 1 ) = 0 <=> x = -24 hoặc x = -1
Vậy ...
Đặt t = \(x^2+14x+24\)
\(\Rightarrow\)\(t\left(t-4x\right)-165x^{^2}=0\)
\(\Leftrightarrow t^2-4xt-165x^2=0\)
\(\Leftrightarrow t^2+11xt-15xt-165x^2=0\)
\(\Leftrightarrow t\left(t+11x\right)-15x\left(t+11x\right)=0\)
\(\Leftrightarrow\left(t+11x\right)\left(t-15x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t+11x=0\\t-15x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=-11x\\t=15x\end{cases}}}\)
với t= -11x
\(\Rightarrow x^2+14x+24=-11x\)
\(\Leftrightarrow x^2+25x+24=0\)
\(\Leftrightarrow x^2+x+24x+24=0\)
\(\Leftrightarrow x\left(x+1\right)+24\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+24\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+24=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-24\end{cases}}}\)
với t=15x
\(\Rightarrow x^2+14x+24=15x\)
\(\Leftrightarrow x^2-x+24=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{95}{4}=0\)(Vô Lí)
vậy....
