b) \(\left(x-7\right)\left(x-4\right)\left(x-5\right)\left(x-2\right)=72\)
\(\Leftrightarrow\) \(\left[\left(x-7\right)\left(x-2\right)\right].\left[\left(x-4\right)\left(x-5\right)\right]\) \(=72\)
\(\Leftrightarrow\) (\(x^2-9x+14\))(\(x^2-9x+20\)) \(=72\) (1)
Đặt \(x^2-9x+17=y\) .Khi đó (1) trở thành:
\(\left(y-3\right)\left(y+3\right)=72\)
\(\Leftrightarrow\) \(y^2-9=72\)
\(\Leftrightarrow\) \(y^2=81\) \(\Leftrightarrow\) \(y\) ∈ \(\left\{9;-9\right\}\)
+)Nếu \(y=9\) \(\Rightarrow\) \(x^2-9x+17=9\)
\(\Leftrightarrow\) \(x^2-9x-8=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\frac{9+\sqrt{113}}{2}\\x=\frac{9-\sqrt{113}}{2}\end{matrix}\right.\)
+)Nếu \(y=-9\) \(\Rightarrow x^2-9x+17=-9\)
\(\Leftrightarrow\) \(x^2-9x+26=0\)
\(\Leftrightarrow\)( \(x^2-2.x.\frac{9}{2}+\left(\frac{9}{2}\right)^2\)) \(+\frac{23}{4}\)\(=0\)
\(\Leftrightarrow\) \(\left(x-\frac{9}{2}\right)^2\)\(=-\frac{23}{4}\)( Vô lí,vì \(\left(x-\frac{9}{2}\right)^2\) ≥0)
Vậy phương trình có tập nghiệm S=\(\left\{\left(\frac{9+\sqrt{113}}{2}\right);\left(\frac{9-\sqrt{113}}{2}\right)\right\}\)
