Form4 binomial , inequality , MI(ans in eng,explain in chin)

Question

binomial:
[4x^2 + 1/(2*開方x) ]^10 find constant term (ANS:45/16,講埋其實constant點找?公式?)

solve inequality :
x lx+3l = -4 (ANS:分左case,唔明)

lx(x-2)l = 1 [ANS: (x-1)^2=0 or 1+- 開方2]

BY MI:
5^2n - 2^4n divisible by9 (show step 3 only,要最快step)

9^n+1 - 8n - 9 divisible by 64 (show step 3 only,要最快step)

Answer 1

[4x^2 + 1/(2*開方x) ]^10 find constant term (ANS:45/16,講埋其實constant點找?公式?)

Constant 即係指 the term independent of x, 換句話說即係 x 的零次方.
所以在你的 expansion 中, 可以假設在 10Cr, 其中 0 <= r <= 10 時, x 的次方為零:
10Cr (4x^2)^r [1/(2*開方x)]^(10-r) is the general term.
In which the power of x is given by:
x^(2r)/[(x^(1/2))^10-r]
= x^(2r)/x^(5-r/2)
=x^(2r- 5 + r/2)
For the power of x to be zero,
2r- 5 + r/2 = 0
r = 2

So the constant term is given by
10C2 (4x^2)^2 [1/(2*開方x)]^(10-2)
=45 (4^2)[1/(2^8)]
=(45 x 16)/256
= 45/16

solve inequality :
x lx+3l = -4 (ANS:分左case,唔明)
分 case 的目的係假設, 呢題應該分兩個 case:
1) x < -3
2) x >= -3
其中 case (1) 時因為假設了 x < -3, 所以 lx+3l = -x-3
變成 -x(x+3) = -4
x^2 + 3x - 4 = 0
x = 1 (rejected, 因為假設了 x < -3) or x = -4 (no need to reject since x < -3)

lx(x-2)l = 1 [ANS: (x-1)^2=0 or 1+- 開方2]
x(x-2) = 1 or x(x-2) = -1 (註: when |y| = k where k is positive or zero, then y = k or -k)
x^2 - 2x - 1 = 0 or x^2 - 2x + 1 = 0
x = 1 + 開方(2) or 1 - 開方(2) or 1

5^2n - 2^4n divisible by9 (show step 3 only,要最快step)
Assume 了 n=k 對的, 做 n=k+1
5^2(k+1) - 2^4(k+1)
= 5^2k x 5^2 - 2^4k x 2^4
= 25 x 5^2k - 16 x 2^4k
= 25(5^2k - 2^4k) + 9 x 2^4k
= 25(9M) + 9 x 2^4k (where M is a natural number)
= 9(25M + 2^4k)
So 5^2(k+1) - 2^4(k+1) is divisible by 9

9^n+1 - 8n - 9 divisible by 64 (show step 3 only,要最快step)
Assume 了 n=k 對的, 做 n=k+1
9^(k+2) - 8(k+1) - 9
= 9 x 9^(k+1) - 8k - 8 - 9
= 9 x [9^(k+1) - 8k - 9] - 8k - 17 + 72k + 81
= 9(64N) + 64k - 64 (where N is a natural number)
= 64(9N + k - 1)
So 9^(k+2) - 8(k+1) - 9 is divisible by 64