Lời giải:
$21^{20}-1=(21-1)(21^{19}+21^{18}+...+21+1)$
$=20(21^{19}+21^{18}+...+21+1)$
Ta thấy:
$21\equiv 1\pmod {10}$ nên:
$21^{19}+21^{18}+...+21+1\equiv 1+1+...+1\equiv 20\equiv 0\pmod {10}$
Hay $21^{19}+21^{18}+...+21+1\vdots 10$
$\Rightarrow 21^{20}-1=20(21^{19}+21^{18}+...+21+1)\vdots 200$
Ta có đpcm.
CMR: 1 - 1/2+ 1/3 -1/4+......+1/99-100 = 1/101 + 1/102+......+1/200
CMR: 1-1/2+1/3-1/4+.....+1/99-1/100 = 1/101 +1/102 +....+ 1/200
\(x^{200}+x^{100}+1⋮x^4+x^2+1\)
How many numbers ranging from 1 to 200 are multiples of 5 and divisible by 3?
Chứng minh rằng:\(x^{200}+x^{100}+1⋮x^4+x^2+1\) với mọi x thuộc Z
Cmr : 1 + 1/1.2 + 1/1.2.3 + .....+ 1/1.2.3....n < 2
CMR \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}< 2\)
1.Cmr 46n+296.13n chia hết cho 1947 với n >0,n thuộc N,n lẻ
2.Cmr 22n(22n+1-1)-1 chia hết cho 9 với n thuộc N*
Cho x+y=1. CMR xy<1/4
cho -1 ≤ a,b,c ≤ 1 va 1 + 2abc ≥ a2 + b2 +c2. cmr: 1 + 2a2b2c2 ≥ a4 + b4 + c4
