Toán Lớp 8: cho a+b=5,ab=-2(a
Toán Lớp 8: cho a+b=5,ab=-2(a
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Toán Lớp 8: cho a+b=5,ab=-2(a
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\left\{ \begin{array}{l}
a + b = 5\\
a.b = – 2
\end{array} \right.\\
+ ){a^2} + {b^2}\\
= {a^2} + 2ab + {b^2} – 2ab\\
= {\left( {a + b} \right)^2} – 2ab\\
= {5^2} – 2.\left( { – 2} \right)\\
= 25 + 4 = 29\\
+ )\dfrac{1}{{{a^3}}} + \dfrac{1}{{{b^3}}}\\
= \dfrac{{{b^3} + {a^3}}}{{{a^3}{b^3}}}\\
= \dfrac{{\left( {a + b} \right)\left( {{a^2} – ab + {b^2}} \right)}}{{{{\left( {ab} \right)}^3}}}\\
= \dfrac{{5.\left[ {{{\left( {a + b} \right)}^2} – 3ab} \right]}}{{{{\left( { – 2} \right)}^3}}}\\
= – \dfrac{{5\left( {{5^2} – 3.\left( { – 2} \right)} \right)}}{8}\\
= – \dfrac{{5.\left( {25 + 6} \right)}}{8}\\
= – \dfrac{{155}}{8}\\
+ )a – b\\
= – \sqrt {{{\left( {a – b} \right)}^2}} \left( {do:a < b} \right)\\
= – \sqrt {{{\left( {a + b} \right)}^2} – 4ab} \\
= – \sqrt {{5^2} – 4.\left( { – 2} \right)} \\
= – \sqrt {33} \\
+ ){a^3} – {b^3}\\
= \left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= – \sqrt {33} .\left[ {{{\left( {a + b} \right)}^2} – ab} \right]\\
= – \sqrt {33} .\left( {{5^2} – \left( { – 2} \right)} \right)\\
= – \sqrt {33} .27\\
= – 27\sqrt {33}
\end{array}$