Bài 2:
`(2/3 - x^2)^5`
Áp dụng tam giác Pascal, ta được:
`= (2/3)^{10} + 5.(2)/(3).x^2 + 10.(2)/(3).x^4 + 10.(2)/(3).x^{6} + 5.(2)/(3).x^8 + x^{10}`
Bài 3:
`(2/(5x^2) - 5/(3)x^3)^{13}`
`= sum_{k = 0}^{13}.C_{13}^{k}.(2/5)^{13 - k}.(-5/3)^{k}.(x^{3k})/(x^{26 - 2k})`
`= sum_{k = 0}^{13}.C_{13}^{k}.(2/5)^{13 - k}.(-5/3)^{k}.x^{5k - 26}`
Giả thuyết
`-> 5k - 26 = 14`
`-> k = 8`
`-> C_{13}^{8}.(2/5)^{5}.(5/3)^{8}.x^{14}`