\(\left(4x-5\right)^2\left(2x-3\right)\left(x-1\right)=9\)
\(\Leftrightarrow\left(4x-5\right)^2\left(2x-3\right).2.\left(x-1\right).4=9.2.4\)
\(\Leftrightarrow\left(4x-5\right)^2\left(4x-6\right)\left(4x-4\right)=72\)(1)
Đặt \(4x-5=a\)
Khi đó (1) trở thành:
\(a^2\left(a-1\right)\left(a+1\right)=72\)
\(\Leftrightarrow a^2\left(a^2-1\right)=72\)
\(\Leftrightarrow a^4-a^2-72=0\)
\(\Leftrightarrow a^4-9a^2+8a^2-72=0\)
\(\Leftrightarrow a^2\left(a^2-9\right)+8\left(a^2-9\right)=0\)
\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0\)
\(\Leftrightarrow a^2-9=0\) (vì \(a^2+8>0\forall a\) )
\(\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)
- Với \(a=3\Rightarrow4x-5=3\Rightarrow x=2\)
-Với \(a=-3\Rightarrow4x-5=-3\Rightarrow x=\frac{1}{2}\)
Vậy \(x=2,x=\frac{1}{2}\)
Chúc bạn học tốt.
