Toán Lớp 8: Nhanh giúp mk
CM:P=a^4-4a^3+5a^2-4a+5>0AAa
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Toán Lớp 8: Nhanh giúp mk
CM:P=a^4-4a^3+5a^2-4a+5>0AAa
Comments ( 2 )
P = {a^4} – 4{a^3} + 5{a^2} – 4a + 5\\
= \left( {{a^4} – 4{a^3} + 4{a^2}} \right) + \left( {{a^2} – 4a + 4} \right) + 1\\
= {a^2}.\left( {{a^2} – 4a + 4} \right) + \left( {{a^2} – 4a + 4} \right) + 1\\
= \left( {{a^2} – 4a + 4} \right)\left( {{a^2} + 1} \right) + 1\\
= \left( {{a^2} – 2.a.2 + {2^2}} \right)\left( {{a^2} + 1} \right) + 1\\
= {\left( {a – 2} \right)^2}.\left( {{a^2} + 1} \right) + 1\\
{\left( {a – 2} \right)^2} \ge 0,\,\,\,\forall a\\
{a^2} + 1 \ge 1 > 0,\,\,\,\forall a\\
\Rightarrow P = {\left( {a – 2} \right)^2}\left( {{a^2} + 1} \right) + 1 \ge 1 > 0,\,\,\,\forall a
\end{array}\)