Toán Lớp 8: phân tích thành nhân tử
1,-12x^3y+75xy^3
2,9xy-4a^2xy
3,2xm^3-2x
4,-4+32a^3b^3
5,5xy-40a^3b^3xy
6,-16a^2bx^3-54a^2b
giải chi tiết giúp mk với ạ
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Comments ( 1 )
1) – 12{x^3}y + 75x{y^3}\\
= 3xy\left( {25{y^2} – 4{x^2}} \right)\\
= 3xy\left( {5y – 2x} \right)\left( {5y + 2x} \right)\\
2)\\
9xy – 4{a^2}xy\\
= xy\left( {9 – 4{a^2}} \right)\\
= xy\left( {3 – 2a} \right)\left( {3 + 2a} \right)\\
3)2x{m^3} – 2x\\
= 2x\left( {{m^3} – 1} \right)\\
= 2x\left( {m – 1} \right)\left( {{m^2} + m + 1} \right)\\
4) – 4 + 32{a^3}{b^3}\\
= 4\left( {8{a^3}{b^3} – 1} \right)\\
= 4\left( {2ab – 1} \right)\left( {4{a^2}{b^2} + 2ab + 1} \right)\\
5)5xy – 40{a^3}{b^3}xy\\
= 5xy\left( {1 – 8{a^3}{b^3}} \right)\\
= 5xy\left( {1 – 2ab} \right)\left( {1 + 2ab + 4{a^2}{b^2}} \right)\\
6) – 16{a^2}b{x^3} – 54{a^2}b\\
= – 2{a^2}b\left( {8{x^3} + 27} \right)\\
= – 2{a^2}b\left( {2x + 3} \right)\left( {4{x^2} – 6x + 9} \right)
\end{array}$