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Không giải phương trình \(3{x^2} + 5x – 6 = 0\) , hãy tính giá trị các biểu thức sau biết x1, x2 là nghiệm của phương trình trên:
a) \(A = (3{x_1} – 2{x_2}) + (3{x_2} – 2{x_1})\)
b)\(B = \dfrac{{{x_1}}}{{{x_2} – 1}} + \dfrac{{{x_2}}}{{{x_1} – 1}}\)
c)\(C = \left| {{x_1} – {x_2}} \right|\)
d)\(D = \dfrac{{{x_1} + 2}}{{{x_1}}} + \dfrac{{{x_2} + 2}}{{{x_2}}}\)
Áp dụng hệ thức Viet cho phương trình bậc hai sau đó thay vào các biểu thức A, B, C, D.
\(\left\{ \begin{array}{l}{x_1} + {x_2} = – \dfrac{b}{a}\\{x_1}.{x_2} = \dfrac{c}{a}\end{array} \right.\)
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Áp dụng hệt thức Viet cho phương trình \(3{x^2} + 5x – 6 = 0\)ta có:
\(\left\{ \begin{array}{l}{x_1} + {x_2} = – \dfrac{5}{3}\\{x_1}.{x_2} = – 2\end{array} \right.\)
a)
\(\begin{array}{l}A = (3{x_1} – 2{x_2}) + (3{x_2} – 2{x_1}) \\= 9{x_1}{x_2} – 6x_1^2 – 6x_2^2 + 4{x_1}{x_2}\\ = 13{x_1}{x_2} – 6\left( {x_1^2 + x_2^2} \right) \\= 13{x_1}{x_2} – 6\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} – 2{x_1}{x_2}} \right]\\ = 13.\left( { – 2} \right) – 6.\left[ {{{\left( {\dfrac{{ – 5}}{3}} \right)}^2} – 2.\left( { – 2} \right)} \right] \\= – 26 – 6.\dfrac{{61}}{9} = – \dfrac{{200}}{3}\end{array}\)
b)
\(\begin{array}{l}B = \dfrac{{{x_1}}}{{{x_2} – 1}} + \dfrac{{{x_2}}}{{{x_1} – 1}}\\ = \dfrac{{{x_1}\left( {{x_1} – 1} \right) + {x_2}\left( {{x_2} – 1} \right)}}{{\left( {{x_2} – 1} \right)\left( {{x_1} – 1} \right)}} \\= \dfrac{{x_1^2 – {x_1} + x_2^2 – {x_2}}}{{{x_1}{x_2} – {x_1} – {x_2}}}\\ = \dfrac{{\left( {x_1^2 + x_2^2} \right) – \left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2} – \left( {{x_1} + {x_2}} \right)}}\\ = \dfrac{{{{\left( {{x_1} + {x_2}} \right)}^2} – 2{x_1}{x_2} – \left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2} – \left( {{x_1} + {x_2}} \right)}}\\ = \dfrac{{{{\left( { – \dfrac{5}{3}} \right)}^2} – 2.\left( { – 2} \right) + \dfrac{5}{3}}}{{ – 2 + \dfrac{5}{3}}} = – \dfrac{{76}}{3}\end{array}\)
c)
\(\begin{array}{l}C = \left| {{x_1} – {x_2}} \right| \\\Leftrightarrow {C^2} = {\left( {{x_1} – {x_2}} \right)^2}\\\;\;\;\;\;\;\;\;\;\; = x_1^2 + x_2^2 – 2{x_1}{x_2}\\ \;\;\;\;\;\;\;\;\;\;= {\left( {{x_1} + {x_2}} \right)^2} – 4{x_1}{x_2} \\\;\;\;\;\;\;\;\;\;\;= {\left( {\dfrac{{ – 5}}{3}} \right)^2} – 4.\left( { – 2} \right) = \dfrac{{97}}{9}\\ \Rightarrow C = \dfrac{{\sqrt {97} }}{3}\end{array}\)
d)
\(\begin{array}{l}D = \dfrac{{{x_1} + 2}}{{{x_1}}} + \dfrac{{{x_2} + 2}}{{{x_2}}} \\\;\;\;\;= \dfrac{{\left( {{x_1} + 2} \right){x_2} + \left( {{x_2} + 2} \right){x_1}}}{{{x_1}{x_2}}} \\\;\;\;\;= \dfrac{{{x_1}{x_2} + 2{x_2} + {x_1}{x_2} + 2{x_1}}}{{{x_1}{x_2}}}\\\;\;\;\; = \dfrac{{2{x_1}{x_2} + 2\left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2}}} \\\;\;\;\;= \dfrac{{2.\left( { – 2} \right) + 2.\left( {\dfrac{{ – 5}}{3}} \right)}}{{ – 2}} = \dfrac{{11}}{3}\end{array}\)