- Ta có : \(A\left( y \right) = 7{y^2} - 3y + \dfrac{1}{2}\)
\(A(5)\) là giá trị của đa thức \(A(y)\) tại \(y = 5\).
\(\eqalign{
& \Rightarrow A\left( 5 \right) = {7.5^2} - 3.5 + {1 \over 2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7.25 - 15 + {1 \over 2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 175 - 15 + {1 \over 2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 160 + {1 \over 2} = {{321} \over 2} \cr} \)
\(B\left( x \right) = 2{x^5} - 3x + 7{x^3} + 4{x^5} + \dfrac{1}{2}\)
Trước hết, ta rút gọn B :
\(\eqalign{& B\left( x \right) = 2{x^5} - 3x + 7{x^3} + 4{x^5} + {1 \over 2} \cr & B\left( x \right) = \left( {2{x^5} + 4{x^5}} \right) - 3x + 7{x^3} + {1 \over 2} \cr & B\left( x \right) = 6{x^5} - 3x + 7{x^3} + {1 \over 2} \cr} \)
\(B(-2)\) là giá trị của đa thức \(B(x)\) tại \(x = -2\).
\(\eqalign{& B\left( { - 2} \right) = 6.{\left( { - 2} \right)^5} - 3.\left( { - 2} \right) + 7.{\left( { - 2} \right)^3} + {1 \over 2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6.( - 32) - ( - 6) + 7.( - 8) + {1 \over 2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 192 + 6 - 56 + {1 \over 2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \left( {192 - 6 + 56} \right) + {1 \over 2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 242 + {1 \over 2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{ - 484} \over 2} + {1 \over 2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{\left( { - 484 + 1} \right)} \over 2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{ - 483} \over 2} \cr} \)
