Toán Lớp 8: x^4+x^3+x^2+1
(2x+1)^2+(x+1)^2
x^4+4x^2+5
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Toán Lớp 8: x^4+x^3+x^2+1
(2x+1)^2+(x+1)^2
x^4+4x^2+5
Phân tích giùm mình với ạ
Comments ( 1 )
a)\\
{x^4} + {x^3} + {x^2} + 1\\
\begin{array}{*{20}{l}}
{ = {\rm{ }}\left( {{x^4}\; – {\rm{ }}{x^3}} \right){\rm{ }} – {\rm{ }}\left( {{x^2}\; – {\rm{ }}1} \right)}\\
{ = {\rm{ }}{x^3}\left( {x{\rm{ }} – {\rm{ }}1} \right){\rm{ }} – {\rm{ }}\left( {x{\rm{ }} – {\rm{ }}1} \right)\left( {x{\rm{ }} + {\rm{ }}1} \right)}
\end{array}\\
= {\rm{ }}\left( {x{\rm{ }} – {\rm{ }}1} \right)({x^3}\; – {\rm{ }}x{\rm{ }} – {\rm{ }}1)\\
b)\\
{(2x + 1)^2} + {(x + 1)^2}\\
= 4{x^2} + 4x + 1 + {x^2} + 2x + 1\\
= 5{x^2} + 6x + 2\\
\Rightarrow sai\,\,de,\,\,sua\,\,de:\\
{(2x + 1)^2} – {(x + 1)^2}\\
= \left( {2x + 1 – x + 1} \right)\left( {2x + 1 + x – 1} \right)\\
= \left( {x + 2} \right)3x\\
c)\\
{x^4} + 4{x^2} + 5\\
= {x^4} + 4{x^2} + 4 – 9\\
= {\left( {{x^2} + 2} \right)^2} – {3^2}\\
= \left( {{x^2} + 2 – 3} \right)\left( {{x^2} + 2 + 3} \right)\\
= \left( {{x^2} – 1} \right)\left( {{x^2} + 5} \right)\\
= \left( {x – 1} \right)\left( {x + 1} \right)\left( {{x^2} + 5} \right)
\end{array}\]