Cho phương trình bậc hai 2x^2 + 5x -1 =0 Không giải phương trình, hãy tính: a) x1+x2 và x1.x2 b) x1^2 + x2^2 c) x1^3 +x2^3 d) x1^2 – x2^2 e) x1^3 – x2

Đáp án:

$\begin{array}{l}
a)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{ – 5}}{2}\\
{x_1}{x_2} =  – \frac{1}{2}
\end{array} \right.\\
b)x_1^2 + x_2^2\\
 = {\left( {{x_1} + {x_2}} \right)^2} – 2{x_1}{x_2}\\
 = {\left( { – \frac{5}{2}} \right)^2} – 2.\left( { – \frac{1}{2}} \right)\\
 = \frac{{25}}{4} + 1 = \frac{{29}}{4}\\
c)x_1^3 + x_2^3\\
 = {\left( {{x_1} + {x_2}} \right)^3} – 3{x_1}{x_2}\left( {{x_1} + {x_2}} \right)\\
 = {\left( { – \frac{5}{2}} \right)^3} – 3.\left( { – \frac{1}{2}} \right).\left( { – \frac{5}{2}} \right)\\
 =  – \frac{{155}}{8}\\
d)x_1^2 – x_2^2\\
 = \left( {{x_1} + {x_2}} \right)\left( {{x_1} – {x_2}} \right)\\
 = \frac{{ – 5}}{2}.\sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} – 4{x_1}{x_2}} \\
 =  – \frac{{5\sqrt {33} }}{4}\\
e)x_1^3 – x_2^3\\
 = \left( {{x_1} – {x_2}} \right)\left( {x_1^2 + {x_1}{x_2} + x_2^2} \right)\\
 = \frac{{\sqrt {33} }}{2}.\left( {\frac{{29}}{4} – \frac{1}{2}} \right)\\
 = \frac{{27\sqrt {33} }}{8}\\
g)x_1^4 – x_2^4\\
 = \left( {x_1^2 + x_2^2} \right)\left( {x_1^2 – x_2^2} \right)\\
 = \frac{{29}}{4}.\left( { – \frac{{5\sqrt {33} }}{4}} \right)\\
 = \frac{{ – 145\sqrt {33} }}{{16}}
\end{array}$