* Mình xin sửa lại đề :
Cho hai đa thức \(f\left(x\right)=2x^2+ax+4\) và \(g\left(x\right)=x^2-5x-b\)( \(a,b\) là hằng số )
Tìm các hệ số a,b sao cho \(f\left(1\right)=g\left(2\right)\)và \(f(-1)=g(5)\)
Bài làm :
\(f\left(x\right)=2x^2+ax+4\)
\(\Rightarrow f\left(1\right)=2\left(1\right)^2+a\left(1\right)+4=2+a+4=6+a\)
\(\Rightarrow f\left(-1\right)=2\left(-1\right)^2+a\left(-1\right)+4=2+\left(-a\right)+4=6-a\)
\(g\left(x\right)=x^2-5x-b\)
\(\Rightarrow g\left(2\right)=\left(2\right)^2-5\left(2\right)-b=4-10-b=-6-b\)
\(\Rightarrow g\left(5\right)=\left(5\right)^2-5\left(5\right)-b=25-25-b=-b\)
Vì \(f\left(1\right)=g\left(2\right)\Rightarrow6+a=-6-b\)
\(f(-1)=g(5)=> 6-a=-b\)
\(\Rightarrow\left\{{}\begin{matrix}6+a=-6-b\\6-a=-b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-12\\-a+b=-6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=-12-a\\-6=-12-a-a\end{matrix}\right.\)
\(\Rightarrow-a-12-a=-6\)
\(\Rightarrow-2a=6\)
\(\Rightarrow a=-3\)
\(\Rightarrow b=6-\left(-3\right)=9\)
Vậy : ......
