32000-4 × 31999+10 × Can 31998 be divided by 7? Try to explain why

32000-4 × 31999+10 × Can 31998 be divided by 7? Try to explain why

Because 32000-4 × 31999+10 × thirty-one thousand nine hundred and ninety-eight
=31998 × （32-4 × 3+10）
=31998 × 7，
So 32000-4 × 31999+10 × 31998 can be divided by seven

32000-4 × 31999+10 × Can 31998 be divided by 7? Try to explain why

Because 32000-4 × 31999+10 × thirty-one thousand nine hundred and ninety-eight
=31998 × （32-4 × 3+10）
=31998 × 7，
So 32000-4 × 31999+10 × 31998 can be divided by seven

The third power of 1999 - the second power of 1999 may be divided by 2000. In 1998, I was the one just now, ha ha

one thousand nine hundred and ninety-nine ³- one thousand nine hundred and ninety-nine ²
=1999 ² (1999-1)
=1999 ²× one thousand nine hundred and ninety-eight
It can be divided by 1998, not 2000

Can the 2000 power of 3 minus four times the 1999 power of 3 plus the 1998 power of 3 be divided by 7, and explain the reasons

2000-4·3~1999+3~1998=9·3~1998-12·3~1998+3~1998
=-2·3~1998
This is obviously not divisible
If you read the title correctly

How to prove that the 99th power of 2 + the 99th power of 3 can be divided by 7?

Binomial expansion is simple if you have learned it
2^99 + 3^99
= 8^33 + 27^33
= (1+7)^33 + (-1 + 28)^33
= (1 + 33·7 + ...+ 7^33) + (-1 + 33·28 - ...+ 28^33)
= (33·7 + ...+ 7^33) + (33·28 - ...+ 28^33)
Each term here contains 7 or 28 and can be divided by 7, so 2 ^ 99 + 3 ^ 99 can be divided by 7
There is another way:
2^99 + 3^99
= 8^33 + 27^33
And x ^ 33 + y ^ 33 can be divided by (x + y) (when n is odd, x ^ n + y ^ n all contain factor X + y)
SO 2 ^ 99 + 3 ^ 99 can be divided by 35, so it can be divided by 7

Given SiNx + cosx = (1-radical 3) / 2, find TaNx? Can you calculate the value of SiNx cosx?

Square both sides of the conditional equation and get it
sinxcosx=-(√3)/4
Binding sin ² x+cos ² X = 1 and TaNx = SiNx / cosx
(tanx)/(1+tan ² x)=-(√3)/4
The solution is: TaNx = - √ 3. Or TaNx = - (√ 3) / 3