`n = 1 =>` Số nguyên tố.

`n ne 1 => n^4 equiv 1(mod 5)`

`=> n^4 + 4 equiv 1 +4 equiv 5 equiv 0 (mod 5)` là hợp số

tưởng pay nick ròi :v 

\(A=\dfrac{19^{20}+5}{19^{20}-8}=\dfrac{19^{20}-8+13}{19^{20}-8}=1+\dfrac{13}{19^{20}-8}\)

\(B=\dfrac{19^{21}-7+13}{19^{21}-7}=1+\dfrac{13}{19^{21}-7}\)

Mà \(19^{21}-7>19^{20}-8\)

=> \(A>B\)

Ta có: 

A = \(\dfrac{10^7+5}{10^7-8}=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)

\(B=\dfrac{10^8+6}{10^8-7}=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)

Mà \(10^8-7>10^7-8\)

=> \(1+\dfrac{13}{10^7-8}>1+\dfrac{13}{10^8-7}\)

=> A < B 

Vậy A < B

Xin lỗi mình kết luận sai vì nhìn nhầm. Đáp án đúng là A > B và cả quá trình trên vẫn đúng nha.

a. 7/421 > 5/531 

b. -4/93 < -3/134

Biện luận trước khi giải: \(a,b\inℕ^∗\). Khi a hoặc b bằng 0 thì biểu thức không xác định.

Bài làm:

Ta có \(1+2+3+...+a=\dfrac{a\left(a+1\right)}{2}\)

Và \(1+2+3+...+b=\dfrac{b\left(b+1\right)}{2}\)

Suy ra \(\dfrac{a\left(a+1\right)}{2a}< \dfrac{b\left(b+1\right)}{2b}\) <=> \(\dfrac{a+1}{2}< \dfrac{b+1}{2}\)

<=> \(a+1< b+1\) <=> a < b

-8x = 32 x 3 => x = 96 : ( -8) = -12 

x^2 = 8 x 2 => x^2 = 16 => x = 4 hoặc x = -4

    \(\dfrac{32}{x}=\dfrac{-8}{3}\)

⇔ \(-8x=96\)

⇔     \(x=-12\)

     \(\dfrac{x}{8}=\dfrac{2}{x}\)

⇔  \(x^2=16\)

⇔  \(x^2=4^2\)

⇔  \(x=4\)

\(\dfrac{5}{6}+\dfrac{-14}{6}\)

\(=\dfrac{5+\left(-14\right)}{6}\)

\(=\dfrac{5-14}{6}\)

\(=\dfrac{-9}{6}\)

\(=\dfrac{-3}{2}\)

\(\dfrac{2}{5}+\dfrac{-7}{4}\)

\(=\dfrac{2\cdot4}{5\cdot4}+\dfrac{-7\cdot5}{4\cdot5}\)

\(=\dfrac{8}{20}+\dfrac{-35}{20}\)

\(=\dfrac{8+\left(-25\right)}{20}\)

\(=\dfrac{-17}{20}\)

-27 phần 20 mà

\(P=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{1225}\right)\left(1-\dfrac{1}{1275}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{1224}{1225}.\dfrac{1274}{1275}\)

\(=\dfrac{2.2}{3.2}.\dfrac{5.2}{6.2}.\dfrac{9.2}{10.2}...\dfrac{1224.2}{1225.2}.\dfrac{1274.2}{1275.2}\)

\(=\dfrac{4}{9}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{2448}{2450}.\dfrac{2548}{2550}\)

\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{48.51}{49.50}.\dfrac{49.52}{50.51}\)

\(=\dfrac{1.2.3...48.49}{2.3.4...49.50}.\dfrac{4.5.6...51.52}{3.4.5...50.51}\)

\(=\dfrac{1}{50}.\dfrac{52}{3}\)

\(=\dfrac{26}{75}\).

\(\dfrac{8}{9}\) < 1 < \(\dfrac{7}{6}\)

Vậy \(\dfrac{8}{9}\) < \(\dfrac{7}{6}\)

7/6 > 1 8/9 < 1

vậy 7/9 > 8/9

HT