F.3數學問題~~(急)－hzb53jl55v的部落格

標題:

F.3數學問題~~(急)

發問:

唔該打埋步驟出黎,thxGiven that n is a positive integer, simplify the following expressions1.4x3^n+2-5x3^n+1 / 10x3^n-8x3^n+12.3^n+1(-3^2)^n / 9^3n+1Given that n is a positive integer when n>2, simplify the following expressions 3.(a^n b^n+1)^2/a^n-1b^n+24.(a^2... 顯示更多唔該打埋步驟出黎,thx Given that n is a positive integer, simplify the following expressions 1.4x3^n+2-5x3^n+1 / 10x3^n-8x3^n+1 2.3^n+1(-3^2)^n / 9^3n+1 Given that n is a positive integer when n>2, simplify the following expressions 3.(a^n b^n+1)^2/a^n-1b^n+2 4.(a^2 b^3)^n(ab)^n+1 5.(x^2n+1+x^2n+2)/x^2n+3+x^2n+1 6.2x^n+3-x^2n / 2^n+3-2x^n+6 Arrange the following numbers in descending order 7. 4^100,8^68,16^49,32^41 8. (-1/8)^40,(1/4)^61,(1/16)^32,1/2^123

最佳解答:

1. [ (4)3^(n+2)-(5)3^(n+1) ] / [ (10)3^n-(8) 3^(n+1) ] = [ (36)3^n- (15)3^n ] / [ (10)3^n-(24) 3^n ] = [ (21)3^n ] / [ (-14)3^n ] = -21/14 = -3/2 2. 睇唔明打乜 Given that n is a positive integer when n>2, simplify the following expressions 3. [a^n b^(n+1)]^2/[ a^(n-1)b^(n+2) ] = [a^(2n) b^(2(n+1)) ] / [ a^(n-1)b^(n+2) ] = [a^(2n) b^(2n+2) ] / [ a^(n-1)b^(n+2) ] = a^(2n-(n-1) ) b^(2n+2- (n+2)) = a^(n+1) b^n 4. (a^2 b^3)^n (ab)^(n+1) = a^(2n) b^(3n) a^(n+1) b^(n+1) = a^(2n+n+1) b^(3n+n+1) = a^(3n+1) b^(4n+1) 5. [x^(2n+1)+x^(2n+2) ]/ [x^(2n+3)+x^(2n+1) ] = [x x^(2n)+ x^2 x^(2n) ]/ [x^3 x^(2n)+ x x^(2n) ] = (x + x^2) x^(2n) / [(x^3 + x) x^(2n) ] = (x + x^2) / (x^3 + x) = x (1 + x) / [ x (x^2 +1) ] = (1 + x) / (x^2 +1) 6. [ 2x^(n+3)-x^(2n) ] / [ 2^(n+3)-2x^(n+6) ] 有無打錯野? 又唔用括號, 邊個分子分母都睇唔清,點做? Arrange the following numbers in descending order 7. 4^100 = (2^2)^100 = 2^200 8^68 = (2^3)^68 = 2^204 16^49 = (2^4)^49 = 2^196 32^41 = (2^5)^41 = 2^205 32^41 > 8^68 > 4^100 > 16^49 8. (-1/8)^40 = 1/8^40 = 1/2^120 (1/4)^61 = 1/4^61 = 1/ 2^122 (1/16)^32 = 1/16^32 = 1/ 2^128 1/2^123 (-1/8)^40 > (1/4)^61 > 1/2^123 > (1/16)^32