by Phân tích đa thức thành nhân tử:
\(a)\ 15{{x}^{2}}+10xy\) \(b)\ 35x\left( y-8 \right)-14y\left( 8-y \right)\)
\(c)\ -x+6{{x}^{2}}y-12xy+2\) \(d)\ {{x}^{3}}-{{x}^{2}}+x-1\)
A. \(\begin{array}{l}
a)\,\,5x\left( {3x + 2y} \right)\\
b)\,\,\left( {5x + 2y} \right)\left( {y – 8} \right)\\
c)\,\,\left( {6xy – 1} \right)\left( {x – 2} \right)\\
d)\,\,\,\left( {{x^2} + 1} \right)\left( {x – 1} \right)
\end{array}\)
B. \(\begin{array}{l}
a)\,\,5x\left( {3x + 2y} \right)\\
b)\,\,7\left( {5x + 2y} \right)\left( {y – 8} \right)\\
c)\,\,\left( {6xy – 1} \right)\left( {x – 2} \right)\\
d)\,\,\,\left( {{x^2} + 1} \right)\left( {x + 1} \right)
\end{array}\)
C. \(\begin{array}{l}
a)\,\,5x\left( {3x + 2y} \right)\\
b)\,\,7\left( {5x + 2y} \right)\left( {y – 8} \right)\\
c)\,\,\left( {6xy – 1} \right)\left( {x – 2} \right)\\
d)\,\,\,\left( {{x^2} + 1} \right)\left( {x – 1} \right)
\end{array}\)
D. \(\begin{array}{l}
a)\,\,5x\left( {3x – 2y} \right)\\
b)\,\,7\left( {5x + 2y} \right)\left( {y – 8} \right)\\
c)\,\,\left( {6xy – 1} \right)\left( {x – 2} \right)\\
d)\,\,\,\left( {{x^2} + 1} \right)\left( {x – 1} \right)
\end{array}\)
Hướng dẫn Chọn đáp án là: C
Lời giải chi tiết:
\(a)\;15{x^2} + 10xy = 5x.3x + 5x.2y = 5x\left( {3x + 2y} \right)\)
\(b)\;35x\left( {y – 8} \right) – 14y\left( {8 – y} \right) \\= 7.5x\left( {y – 8} \right) + 7.2y\left( {y – 8} \right) \\= \left( {7.5x + 7.2y} \right)\left( {y – 8} \right)\\ = 7\left( {5x + 2y} \right)\left( {y – 8} \right)\)
\(\begin{array}{l}c) – x + 6{x^2}y – 12xy + 2\\ = \left( {6{x^2}y – 12xy} \right) – \left( {x – 2} \right)\\ = \left( {6xy.x – 6xy.2} \right) – \left( {x – 2} \right)\\ = 6xy\left( {x – 2} \right) – \left( {x – 2} \right)\\ = \left( {6xy – 1} \right)\left( {x – 2} \right)\end{array}\)
\(\begin{array}{l}d)\;{x^3} – {x^2} + x – 1\\ = {x^2}.x – {x^2} + x – 1\\ = {x^2}\left( {x – 1} \right) + \left( {x – 1} \right)\\ = \left( {{x^2} + 1} \right)\left( {x – 1} \right)\end{array}\)
