$\displaystyle \begin{array}{{>{\displaystyle}l}} y=x^{2} +x+1\\ a) \ Tính\ A=f( 1) .f( 2) -3f\left(\frac{1}{2}\right)\\ f( 1) =1+1+1=3\\ f( 2) =2^{2} +2+1=7\\ f\left(\frac{1}{2}\right) =\left(\frac{1}{2}\right)^{2} +\frac{1}{2} +1\\ =\frac{1}{4} +\frac{2}{4} +\frac{4}{4} =\frac{7}{4} \ \\ Do\ đó\ :\ A=1.7-3.\frac{7}{4} \ \\ A=\frac{28-21}{4} =\frac{27}{4} \ \\ b) \ Tìm\ x\ biết\ f( x) =3;\ f( x) =7\\ +Với\ f( x) =3\ thì\ :\ \\ 3=x^{2} +x+1\\ \rightarrow x^{2} +x-2=0\\ \rightarrow x^{2} +2x-x-2=0\\ \rightarrow x( x+2) -( x+2) =0\\ \rightarrow ( x-1)( x+2) =0\\ \rightarrow \left[ \begin{array}{l l} x-1=0 & \\ x+2=0 & \end{array} \right.\rightarrow \left[ \begin{array}{l l} x=1 & \\ x=-2 & \end{array} \right.\\ +Với\ f( x) =7\ thù\ \\ 7=x^{2} +x+1\\ \rightarrow x^{2} +x-6=0\\ \rightarrow x^{2} +3x-2x-6=0\\ \rightarrow x( x+3) -2( x+3) =0\\ \rightarrow ( x+3)( x-2) =0\ \\ \rightarrow \left[ \begin{array}{l l} x+3=0 & \\ x-2=0 & \end{array} \right.\rightarrow \left[ \begin{array}{l l} x=-3 & \\ x=2\ & \end{array} \right. \ \\ \end{array}$
