Toán Lớp 8: Phân tích các đa thức sau thành nhân tử :
1, x^3 – 2x^2 + 5x
2, 3x(x-1) + 5(x-1)
3, 2x^2(x+1)+4(x+1)
4, -3x-6xy+9xz
5, 2x ^ 2 y – 4xy^2+6xy
6, 4x^3y^2-8y^2y^3+2x^2y
7, 9x^2y^3-3x^4y^2-6x^3y^2+18xy^4
8, 7x^2y^2-21xy^2z+7xyz-14xy
9, a^3x^2y-5/2a^3x^4+3/2a^4x^2y
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1){x^3} – 2{x^2} + 5x\\
= x\left( {{x^2} – 2x + 5} \right)\\
2)3x\left( {x – 1} \right) + 5\left( {x – 1} \right)\\
= \left( {x – 1} \right)\left( {3x + 5} \right)\\
3)2{x^2}\left( {x + 1} \right) + 4\left( {x + 1} \right)\\
= 2.\left( {x + 1} \right)\left( {{x^2} + 2} \right)\\
4) – 3x – 6xy + 9xz\\
= 3x\left( {3z – 2y – 1} \right)\\
5)2{x^2}y – 4x{y^2} + 6xy\\
= 2xy\left( {x – 2y + 3} \right)\\
6)4{x^3}{y^2} – 8{x^2}{y^3} + 2{x^2}y\\
= 2{x^2}y\left( {2xy – 4y + 1} \right)\\
7)9{x^2}{y^3} – 3{x^4}{y^2} – 6{x^3}{y^2} + 18x{y^4}\\
= 3x{y^2}\left( {3xy – {x^3} – 2{x^2} + 6{y^2}} \right)\\
8)\\
7{x^2}{y^2} – 21x{y^2}z + 7xyz – 14xy\\
= 7xy\left( {xy – 3yz + z – 2} \right)\\
9){a^3}{x^2}y – \dfrac{5}{2}{a^3}{x^4} + \dfrac{3}{2}{a^4}{x^2}y\\
= {a^3}{x^2}\left( {y – \dfrac{5}{2}{x^2} + \dfrac{3}{2}ay} \right)
\end{array}$