Sửa đề xíu nha: \(g\left(x\right)=x^2-4.\left(x-2\right)\)
a, Ta có:
\(f\left(2\right)+g\left(\dfrac{-1}{2}\right)=\left(2^2-6.2+4\right)+\left[\left(\dfrac{-1}{2}\right)^2-4.\left(\dfrac{-1}{2}-2\right)\right]\)
\(=\left(4-12+4\right)+\left(\dfrac{1}{4}-4.\dfrac{-5}{2}\right)=-4+\left(\dfrac{1}{4}+10\right)\)
\(=-4+\dfrac{1}{4}+10=6+\dfrac{1}{4}=\dfrac{25}{4}\)
Vậy \(f\left(2\right)+g\left(\dfrac{-1}{2}\right)=\dfrac{25}{4}\)
b, Ta có: \(g\left(x\right)=x^2-4\left(x-2\right)=x^2-4x+8\)
Do đó:
\(f\left(x\right)-g\left(x\right)=\left(x^2-6x+4\right)-\left(x^2-4x+8\right)=x^2-6x+4-x^2+4x-8\)
\(=\left(x^2-x^2\right)-\left(6x-4x\right)+\left(4-8\right)=-2x-4\)
Vậy \(f\left(x\right)-g\left(x\right)=-2x-4\)
c, Ta có:\(h\left(x\right)=\left|f\left(x\right)-g\left(x\right)\right|=\left|-2x-5\right|\)
Mà \(\left|-2x-4\right|\ge0\) với mọi x
Do đó \(h\left(x\right)=\left|-2x-4\right|\) có giá trị nguyên khác 0 và không lớn hơn 1
\(\Leftrightarrow\left|-2x-4\right|=1\Rightarrow\left[{}\begin{matrix}-2x-4=1\\-2x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-2x=5\\-2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2,5\\x=-1,5\end{matrix}\right.\)
Vậy x=-2,5 hoặc x=-1,5.
