6
\(\left(x+2\right)\left(x-3\right)\left(x^2+2x-24\right)=16x^2\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x+6\right)\left(x-4\right)=16x^2\\ \Leftrightarrow\left(x^2+8x+12\right)\left(x^2-7x+12\right)=16x^2\)
Nhận thấy x = 0 không phải là nghiệm phương trình, ta chia cả hai vế phương trình cho x2 ta được
\(\left(x+8+\dfrac{12}{x}\right)\left(x-7+\dfrac{12}{x}\right)=16\)
Đặt \(x+\dfrac{12}{x}=a\) , phương trình trở thành
\(\left(a+8\right)\left(a-7\right)=16\\ \Leftrightarrow a^2+a-72=0\\ \Leftrightarrow\left(a-8\right)\left(a+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=8\\a=-9\end{matrix}\right.\)
TH1 :
\(x+\dfrac{12}{x}=8\\ \Rightarrow x^2-8x+12=0\\ \Leftrightarrow\left(x-6\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
TH2:
\(x+\dfrac{12}{x}=-9\\ \Rightarrow x^2+9x+12=0\\ \Leftrightarrow\left(x+\dfrac{9}{2}\right)^2=\dfrac{33}{4}\\ \Leftrightarrow x+\dfrac{9}{2}=\dfrac{\pm\sqrt{33}}{2}\\ \Leftrightarrow x=\dfrac{-9\pm\sqrt{33}}{2}\)
6) Ta có: \(\left(x+2\right)\left(x-3\right)\left(x^2+2x-24\right)=16x^2\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x+6\right)\left(x-4\right)=16x^2\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)\left(x-3\right)\left(x-4\right)-16x^2=0\)
\(\Leftrightarrow\left(x^2+8x+12\right)\left(x^2-7x+12\right)-16x^2=0\)
\(\Leftrightarrow\left(x^2+12\right)^2+x\left(x^2+12\right)-72x^2=0\)
\(\Leftrightarrow\left(x^2+12\right)^2+9x\left(x^2+12\right)-8x\left(x^2+12\right)-72x^2=0\)
\(\Leftrightarrow\left(x^2+12\right)\left(x^2+9x+12\right)-8x\left(x^2+9x+12\right)=0\)
\(\Leftrightarrow\left(x^2-8x+12\right)\left(x^2+9x+12\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\left(x^2+9x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\\x=\dfrac{-9-\sqrt{33}}{2}\\x=\dfrac{-9+\sqrt{33}}{2}\end{matrix}\right.\)
7b:
Đặt \(\dfrac{x-1}{x+2}=a;\dfrac{x+1}{x-2}=b\left(a,b\ge0\right)\)
PT <=> \(a^2-4ab+3b^2=0\)
<=> \(a^2-ab-3ab+3b^2=0\)
<=> \(a\left(a-b\right)-3b\left(a-b\right)=0\)
<=> (a - b).(a - 3b) = 0
<=> a = b hoặc a = 3b
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}=\dfrac{x+1}{x-2}\\\dfrac{x-1}{x+2}=3.\dfrac{x+1}{x-2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-2\right)=\left(x+1\right)\left(x+2\right)\\\left(x-1\right)\left(x-2\right)=\left(x+2\right).3.\left(x+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=x^2+3x+2\\x^2-3x+2=3x^2+9x+6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-6x=0\\2x^2+12x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+6x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+3\right)^2-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+3+\sqrt{7}\right)\left(x+3-\sqrt{7}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\mp\sqrt{7}\end{matrix}\right.\)
