Ta có: \(\left(x-7\right)\left(x^2-9x+20\right)\left(x-2\right)=72\)
\(\Leftrightarrow\left(x^2-9x+20\right)\left(x^2-9x+14\right)=72\)
Đặt \(x^2-9x+17=a\) khi đó:
\(PT\Leftrightarrow\left(a+3\right)\left(a-3\right)=72\)
\(\Leftrightarrow a^2-9-72=0\)
\(\Leftrightarrow a^2=81\Rightarrow\orbr{\begin{cases}a=9\\a=-9\end{cases}}\)
Nếu a = 9 khi đó \(x^2-9x+17=9\)
\(\Leftrightarrow x^2-9x+8=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=8\end{cases}}\)
Nếu a = -9 khi đó \(x^2-9x+17=-9\)
\(\Leftrightarrow x^2-9x+26=0\)
\(\Leftrightarrow\left(x^2-9x+\frac{81}{4}\right)+\frac{23}{4}=0\)
\(\Leftrightarrow\left(x-\frac{9}{2}\right)^2=-\frac{23}{4}\left(ktm\right)\)
Vậy \(S=\left\{1;8\right\}\)
( x - 7 )( x2 - 9x + 20 )( x - 2 ) = 72
⇔ [ ( x - 7 )( x - 2 ) ]( x2 - 9x + 20 ) - 72 = 0
⇔ ( x2 - 9x + 14 )( x2 - 9x + 20 ) - 72 = 0
Đặt t = x2 - 9x + 17
⇔ ( t - 3 )( t + 3 ) - 72
⇔ t2 - 9 - 72 = 0
⇔ t2 - 81 = 0
⇔ ( t - 9 )( t + 9 ) = 0
⇔ ( x2 - 9x + 17 - 9 )( x2 - 9x + 17 + 9 ) = 0
⇔ ( x2 - 9x + 8 )( x2 - 9x + 26 ) = 0
⇔ ( x2 - 8x - x + 8 )( x2 - 9x + 26 ) = 0
⇔ [ x( x - 8 ) - ( x - 8 ) ]( x2 - 9x + 26 ) = 0
⇔ ( x - 8 )( x - 1 )( x2 - 9x + 26 ) = 0
⇔ x - 8 = 0 hoặc x - 1 = 0 hoặc x2 - 9x + 26 = 0
⇔ x = 8 hoặc x = 1 [ x2 - 9x + 26 = ( x2 - 9x + 81/4 ) + 23/4 = ( x - 9/2 )2 + 23/4 ≥ 23/4 > 0 ∀ x ]
\(\left(x-7\right)\left(x^2-9x+20\right)\left(x-2\right)=72\)
\(\Leftrightarrow\left(x-7\right)\left(x-2\right)\left(x^2-9x+20\right)-72=0\)
\(\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\)(1)
Đặt \(x^2-9x+17=t\)
Thay \(t=x^2-9x+17\)vào (1) ta được:
\(\left(t-3\right)\left(t+3\right)-72=0\)
\(\Leftrightarrow t^2-9-72=0\)\(\Leftrightarrow t^2-81=0\)
\(\Leftrightarrow\left(t-9\right)\left(t+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t-9=0\\t+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=9\\t=-9\end{cases}}\)
TH1: Nếu \(t=9\)\(\Rightarrow x^2-9x+17=9\)
\(\Leftrightarrow x^2-9x+8=0\)\(\Leftrightarrow\left(x^2-x\right)-\left(8x-8\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=8\end{cases}}\)
TH2: Nếu \(t=-9\)\(\Rightarrow x^2-9x+17=-9\)
\(\Leftrightarrow x^2-9x+26=0\)
\(\Leftrightarrow x^2-2.\frac{9}{2}.x+\frac{81}{4}+\frac{23}{4}=0\)
\(\Leftrightarrow\left(x-\frac{9}{2}\right)^2+\frac{23}{4}=0\)
Vì \(\left(x-\frac{9}{2}\right)\ge0\forall x\)\(\Rightarrow\left(x-\frac{9}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}\)
\(\Rightarrow\)Với \(t=-9\)thì phương trình vô nghiệm
Vậy \(x=1\)hoặc \(x=8\)
