\(\left(4x-5\right)^2\left(2x-3\right)\left(x-1\right)=9\)
\(\Leftrightarrow\left(16x^2-40x+25\right)\left(2x^2-3x-2x+3\right)=9\)
\(\Leftrightarrow\left(16x^2-40x+25\right)\left(2x^2-5x+3\right)=9\)
\(\Leftrightarrow\left(16x^2-40x+25\right).8\left(2x^2-5x+3\right)=72\)
\(\Leftrightarrow\left(16x^2-40x+25\right)\left(16x^2-40x+24\right)=72\)
\(\Leftrightarrow\left(16x^2-40x+\dfrac{49}{2}+\dfrac{1}{2}\right)\left(16x^2-40x+\dfrac{49}{2}-\dfrac{1}{2}\right)=72\)
\(\Leftrightarrow\left(16x^2-40x+\dfrac{49}{2}\right)^2-\dfrac{1}{4}=72\)
\(\Leftrightarrow\left(16x^2-40x+\dfrac{49}{2}\right)^2-\dfrac{289}{4}=0\)
\(\Leftrightarrow\left(16x^2-40x+\dfrac{49}{2}-\dfrac{17}{2}\right)\left(16x^2-40x+\dfrac{49}{2}+\dfrac{17}{2}\right)=0\)
\(\Leftrightarrow\left(16x^2-40x+16\right)\left(16x^2-40x+33\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}16x^2-40x+16=0\\16x^2-40x+33=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x-5\right)^2=9\\\left(4x-5\right)^2=-8\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=3\\4x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
\(\left(4x-5\right)^2.\left(2x-3\right).\left(x-1\right)=9\)
\(\Leftrightarrow\left(4x-5\right)^2.\left(4x-6\right).\left(4x-4\right)=72\) (1)
Đặt 4x-5=t
=> phương trình (1) \(\Leftrightarrow\left(t-1\right)\left(t+1\right)t^2=72\)
\(\Leftrightarrow t^4-t^2=72\)
\(\Leftrightarrow t^4-t^2-72=0\)
\(\Leftrightarrow t^4-9t^2+8t^2-72=0\)
\(\Leftrightarrow t^2\left(t^2-9\right)+8\left(t^2-9\right)=0\)
\(\Leftrightarrow\left(t-3\right)\left(t+3\right)\left(t^2+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-3=0\\t+3=0\\t^2+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-5-3=0\\4x-5+3=0\\t^2=-8\left(v\text{ô}l\text{ý}\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}4x=8\\4x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)