*toán tìm x;y->...

N

ngtungson91

H

hiensau99

1. a, ta có: [TEX] 3^y \not\vdots 2; 80 \vdots 2[/TEX] \Rightarrow [TEX]2^x \vdots 2[/TEX]
Leftrightarrow [TEX]2^x=1 \Leftrightarrow x=0[/TEX]. khi đó: [TEX]1+80=81 = 3^y \Leftrightarrow y=4 [/TEX]
b,Ta có:
[TEX] 4x \vdots 2; 399 \not\vdots 2 \Rightarrow 4x+399 \not\vdots 2 [/TEX]. mà [TEX]10^y \not\vdots 2 [/TEX] \Leftrightarrow y=0. Khi đó:
[TEX]4x-399=1 \Rightarrow 4x=400 \Rightarrow x=100[/TEX]

2a) 1972^33-12^5

- Ta có: [TEX]1972^{33} \vdots 2; 12^5 \vdots 2 \Rightarrow 1972^{33}-12^5 \vdots 2[/TEX] (1)
* Xét: [TEX]1972^{33} - 12^5 \equiv 2^{33}-2^5 \equiv 2^{2.16+1}-32 \equiv 4^{16}.2-2 \equiv (-1)^{16}.2-2=0[/TEX] (mod 5) => đpcm (2)
*Ta có: (2,5)=1; 2.5=10 (3)
* từ (1);(2);(3) => chia hết cho 10

b) Nhầm đề chăng?
Có [TEX]1999^{3467}-13^{16} \equiv (-1)^{3467}-3^{16} \equiv -1- 9^8 \equiv -1-(-1)^8=-1-1=-2 [/TEX] (mod 10)
 
Top Bottom