Bài 1:
Ta có [tex]C^p_{m+n}=C^0_nC^p_m+C^1_nC^{p-1}_m+...+C^p_nC^0_m[/tex]
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Đây
Với p=m=n,ta được:
Bài 2:
Ta có:
[TEX]\text{(2+x+\frac{5}{x})^8 = \sum_{i=0}^8.C^i_8. 2^{8-i}(x+\frac{5}{x})^i=\sum_{i=0}^8 C^i_8.2^{8-i}\sum_{k=0}^i.C_i^kx^{i-k}.5^k(\frac{1}{x})^k = \sum_{i=0}^8 C^i_8.2^{8-i}\sum_{k=0}^i.C_i^kx^{i-2k}.5^k\\ So hang tong quat co dang :\\ T_{i,k}= C^i_8.2^{8-i}C_i^kx^{i-2k}.5^k khong chua x \Leftrightarrow i-2k=0 voi 0\leq k\leq i\leq 8, i,k \in N. \\ Cac cap so thoa man dieu kien tren gom: (i,k)=(0,0); (2,1); (4,2).\\ Vay he so cua so hang khong chua x:\\ T= C^0_8.2^8.5^0+C^2_8.2^6.C^1_2.5^1+ C^4_8.2^4.C^2_4.5^2= 186176[/TEX]
OK nhé!