Hệ số tỉ lệ k là \(k=\dfrac{y}{x}=\dfrac{8}{17}\)

Để \(\dfrac{3-x}{5}>0\) thì 3-x>0

=>x<3

=>\(x\in\left\{...;1;2;3\right\}\)

1) \(\dfrac{4^2}{2^3}=\dfrac{\left(2^2\right)^2}{2^3}=\dfrac{2^4}{2^3}=2\)

2) \(\dfrac{25^5}{125^3}=\dfrac{\left(5^2\right)^5}{\left(5^3\right)^3}=\dfrac{5^{10}}{5^9}=5\)

3) \(\dfrac{27^6}{9^9}=\dfrac{\left(3^3\right)^6}{\left(3^2\right)^9}=\dfrac{3^{18}}{3^{18}}=1\) 

4) \(\dfrac{16^{13}}{32^{10}}=\dfrac{\left(2^4\right)^{13}}{\left(2^5\right)^{10}}=\dfrac{2^{52}}{2^{50}}=2^2-4\)

5) \(\dfrac{16^5}{64^4}=\dfrac{\left(4^2\right)^5}{\left(4^3\right)^4}=\dfrac{4^{10}}{4^{12}}=\dfrac{1}{4^2}=\dfrac{1}{16}\)

6) \(\dfrac{81^8}{27^{11}}=\dfrac{\left(3^4\right)^8}{\left(3^3\right)^{11}}=\dfrac{3^{32}}{3^{33}}=\dfrac{1}{3}\)

7) \(\dfrac{6^3}{2^3}=\dfrac{2^3\cdot3^3}{2^3}=3^3=27\)

8) \(\dfrac{5^4}{15^3}=\dfrac{5^4}{3^3\cdot5^3}=\dfrac{5}{3^3}=\dfrac{5}{27}\)

9) \(\dfrac{7^{15}}{14^{13}}=\dfrac{7^{15}}{7^{13}\cdot2^{13}}=\dfrac{7^2}{2^{13}}=\dfrac{49}{2^{13}}\)

10) \(\dfrac{\left(-2\right)^6}{24^2}=\dfrac{2^6}{8^2\cdot3^2}=\dfrac{2^6}{\left(2^3\right)^2\cdot3^2}=\dfrac{2^6}{2^6\cdot3^2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)

11: \(\dfrac{27^2}{\left(-18\right)^3}=\dfrac{-3^6}{\left(3^2\cdot2\right)^3}=\dfrac{-3^6}{3^6\cdot2^3}=\dfrac{-1}{8}\)

12: \(\dfrac{\left(-10\right)^8}{8^3\cdot25^4}=\dfrac{2^8\cdot5^8}{2^6\cdot5^8}=2^2=4\)

13: \(\dfrac{4^4\cdot8^3}{16^4}=\dfrac{2^8\cdot2^9}{2^{16}}=2\)

14: \(\dfrac{5^7\cdot9^2}{15^5}=\dfrac{5^7\cdot3^4}{5^5\cdot3^5}=\dfrac{5^2}{3}=\dfrac{25}{3}\)

15: \(\dfrac{21^{13}}{49^6\cdot\left(-27\right)^4}=\dfrac{-7^{13}\cdot3^{13}}{7^{12}\cdot3^{12}}=-7\cdot3=-21\)

16: \(\dfrac{\left(-18\right)^{21}\cdot27^4}{81^{13}\cdot16^5}=\dfrac{-3^{42}\cdot2^{21}\cdot3^{12}}{3^{52}\cdot2^{20}}=\dfrac{-3^{54}}{3^{52}}\cdot2=-3^2\cdot2=-18\)

17: \(\dfrac{45^{14}\cdot8^2}{6^5\cdot125^4\cdot81^6}=\dfrac{3^{28}\cdot5^{14}\cdot2^6}{2^5\cdot3^5\cdot3^{24}\cdot5^{12}}=\dfrac{3^{28}}{3^{29}}\cdot\dfrac{5^{14}}{5^{12}}\cdot\dfrac{2^6}{2^5}=\dfrac{5^2\cdot2}{3}=\dfrac{50}{3}\)

18: \(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\dfrac{3^{29}\left(11-3\right)}{3^{28}\cdot2^2}=3\cdot\dfrac{8}{4}=3\cdot2=6\)

19: \(\dfrac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{16^4\cdot5^7+20^8}\)

\(=\dfrac{2^{15}\cdot5^8-2^{14}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}=\dfrac{2^{14}\cdot5^8\cdot\left(2-5\right)}{2^{16}\cdot5^7\cdot\left(1+5\right)}=\dfrac{1}{4}\cdot5\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)

1) \(\dfrac{8^{10}}{4^8}=\dfrac{\left(2^3\right)^{10}}{\left(2^2\right)^8}=\dfrac{2^{30}}{2^{16}}=2^{30-16}=3^{14}\)

2) \(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{4^{2+3}}{\left(2^2\right)^5}=\dfrac{4^5}{4^5}=1\)

3) \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{\left(2^3\right)^2\cdot\left(2^2\right)^5}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{2^{16}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)

4) \(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{2^5\cdot3^5\cdot\left(2^3\right)^2}=\dfrac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{2^7\cdot3}{2^{11}}=\dfrac{3}{2^4}=\dfrac{3}{16}\) 

5) \(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{3^6\cdot2^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{3^6\cdot2^6\cdot2^9}=\dfrac{2^{15}\cdot3^2}{2^{15}}=3^2=9\)

6) \(\dfrac{2^7\cdot9^3}{6^3\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{2^3\cdot3^3\cdot\left(2^3\right)^2}=\dfrac{2^7\cdot3^6}{2^3\cdot3^3\cdot2^6}=\dfrac{2^7\cdot3^3}{2^9}=\dfrac{3^3}{2^2}=\dfrac{27}{4}\)

\(1,\left(1\right)\cdot x=1\)

\(\Rightarrow x=\dfrac{1}{1,\left(1\right)}\)

\(\Rightarrow x=1:\dfrac{10}{9}\)

\(\Rightarrow x=\dfrac{9}{10}=0,9\)

Vậy số thập phân x  thỏa mãn là 0,9 

\(1,\left(1\right).x=1\)

\(\left(1+\dfrac{1}{9}\right).x=1\)

\(\dfrac{10}{9}.x=1\)

\(x=1:\dfrac{10}{9}\)

\(x=\dfrac{9}{10}\)

\(x=0,9\)

Yêu cầu đề bài là gì, bạn cần ghi rõ ra nhé.

\(-\dfrac{1}{1\times4}-\dfrac{1}{4\times7}-\dfrac{1}{7\times10}-\dfrac{1}{10\times13}\)

\(=-\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+\dfrac{1}{10\times13}\right)\)

\(=-\dfrac{1}{3}\cdot\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}\right)\)

\(=-\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)\)

\(=-\dfrac{1}{3}\cdot\left(1-\dfrac{1}{13}\right)\)

\(=\dfrac{-1}{3}\cdot\dfrac{12}{13}\)

\(=\dfrac{-4}{13}\)

Thầy ơi thầy xem lại bài ạ 

Ta có:

2024 ≡ 2024 (mod 2025)

2024² ≡ 1 (mod 2025)

2024³ ≡ 2024.2024² (mod 2025) ≡ 2024.1 (mod 2025) ≡ 2024 (mod 2025)

2024⁴ ≡ (2024²)² (mod 2025) ≡ 1² (mod 2025) ≡ 1 (mod 2025)

⇒ 2024 + 2024² + 2024³ + 2024⁴ ≡ 2024 + 1 + 2024 + 1 (mod 2025) 4050 ≡ (mod 2025) ≡ 0 (mod 2025)

Vậy (2024 + 2024² + 2024³ + 2024⁴) ⋮ 2025

\(\Rightarrow2025\cdot2024\cdot\left(1+2024^2\right)\) ⋮ 2025 

\(\Rightarrow\left(2024+2024^2+2024^3+2024^4\right)⋮2025\)

3x=4y=5z

=>\(\dfrac{3x}{60}=\dfrac{4y}{60}=\dfrac{5z}{60}\)

=>\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}\)

mà x-y-z=-21

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y-z}{20-15-12}=\dfrac{-21}{-7}=3\)

=>\(x=3\cdot20=60;y=15\cdot3=45;z=12\cdot3=36\)