Toán Lớp 8: Phân tích các đa thức sau thành phân tử.
A. 2xy + 3z + 6y + xz
B. 16x² – ( x + 1 )²
C. x² – 6x – 7
D. x³ – 2x² + 2x – 1
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Toán Lớp 8: Phân tích các đa thức sau thành phân tử.
A. 2xy + 3z + 6y + xz
B. 16x² – ( x + 1 )²
C. x² – 6x – 7
D. x³ – 2x² + 2x – 1
Comments ( 2 )
{\rm{a) }}2xy{\rm{ }} + {\rm{ }}3z{\rm{ }} + {\rm{ }}6y{\rm{ }} + {\rm{ }}xz\\
= {\rm{ }}\left( {2xy{\rm{ }} + {\rm{ }}6y} \right){\rm{ }} + {\rm{ }}\left( {3z{\rm{ }} + {\rm{ }}xz} \right)\\
= {\rm{ }}2y\left( {x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} + {\rm{ }}z\left( {x{\rm{ }} + {\rm{ }}3} \right)\\
= {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}3} \right)\left( {2y{\rm{ }} + {\rm{ }}z} \right)\\
b)\,16x{\rm{ }} – {\rm{ }}\left( {{\rm{ }}x{\rm{ }} + {\rm{ }}1{\rm{ }}} \right)\\
= {\left( {4x} \right)^2} – {\rm{ }}\left( {{\rm{ }}x{\rm{ }} + {\rm{ }}1{\rm{ }}} \right)\\
= \left( {4x – x – 1} \right)\left( {4x + x – 1} \right)\\
= \left( {3x – 1} \right)\left( {5x – 1} \right)\\
c)\,x{\rm{ }} – {\rm{ }}6x{\rm{ }} – {\rm{ }}7\\
= {x^2} + x – 7x – 7\\
= x(x + 1) – 7(x + 1)\\
= (x + 1)(x – 7)\\
d)x{\rm{ }} – {\rm{ }}2x{\rm{ }} + {\rm{ }}2x{\rm{ }} – {\rm{ }}1\\
= \left( {{x^3} – 1} \right) – \left( {2{x^2} – 2x} \right)\\
= \left( {x – 1} \right)\left( {{x^2} + x + 1} \right) – 2x\left( {x – 1} \right)\\
= \left( {x – 1} \right)\left( {{x^2} + x + 1 – 2x} \right)\\
= \left( {x – 1} \right)\left( {x – x + 1} \right)
\end{array}\]