(x-3)2-4x+20=(4-x)2
x2-6x+9-4x+20=16-8x+x2
-10x+29=16-8x
-10x+8x=16-29
-2x=-13
x=13/2
Ta có : \(\left(x-3\right)^2-4\left(x-5\right)=\left(4-x\right)^2\)
=> \(x^2-6x+9-4x+20=16-8x+x^2\)
=> \(x^2-6x+9-4x+20-16+8x-x^2=0\)
=> \(13-2x=0\)
=> \(x=\frac{13}{2}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{13}{2}\right\}\)
\(\left(x-3\right)^2-4\cdot\left(x-5\right)=\left(4-x\right)^2\)
\(\Leftrightarrow\left(x^2-6x+9\right)-\left(4x-20\right)-\left(16-8x+x^2\right)=0\)
\(\Leftrightarrow x^2-6x+9-4x+20-16+8x-x^2=0\)
\(\Leftrightarrow-2x+13=0\)
\(\Rightarrow x=\frac{13}{2}\)
