\(A=18^{15}-\frac{8^5.9^{14}.5}{4^8}.27^{10}-4^7.3^{31}=\left(2.3^2\right)^{15}-\frac{\left(2^3\right)^5.\left(3^2\right)^{14}.5}{\left(2^2\right)^8}.\left(3^3\right)^{10}-\left(2^2\right)^7.3^{31}\)

\(=2^{15}.3^{30}-\frac{2^{15}.3^{28}.5}{2^{16}}.3^{30}-2^{14}.3^{31}\)

\(=2^{15}.3^{30}-\frac{3^{58}.5}{2}-2^{14}.3^{31}\)

\(=2^{14}.3^{30}\left(2-3\right)-\frac{3^{58}.5}{2}=-1.2^{14}.3^{30}-\frac{3^{58}.5}{2}=-1.2^{15}.3^{30}-3^{58}.5\)

\(=3^{30}\left(-1.2^{15}-3^{28}.5\right)\)

#)Giải :

Ta có :

\(\frac{a}{a'}+\frac{b'}{b}=1\Leftrightarrow ab+a'b'=a'b\Leftrightarrow abc+a'b'c'=a'bc\left(1\right)\)(vì c khác 0)

\(\frac{b}{b'}=\frac{c'}{c}=1\Leftrightarrow bc+b'c'=b'c=\Leftrightarrow a'bc+a'b'c'=a'b'c\left(2\right)\)(vì a' khác 0)

Từ \(\left(1\right)\left(2\right)\Rightarrowđpcm\)

a) 54⁴ và 14⁸    

14⁸ = ( 14² )⁴ = 196⁴

=> 54⁴ < 196⁴

=>54⁴ < 14⁸    

\(a)54^4< 196^4=\left(14^2\right)^4=14^8\)

\(\Rightarrow54^4< 14^8\)

\(b)27^5=\left(3^3\right)^5=\left(3^5\right)^3=143^3< 245^3\)

\(\Rightarrow245^3>27^5\)

\(b)3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}=\left(2^3\right)^{100}=2^{300}\)

\(\Rightarrow3^{200}>2^{300}\)

a, 4n - 7 ⋮ n - 1

=> 4n - 4 - 3 ⋮ n - 1

=> 4(n - 1) - 3 ⋮ n - 1

=> -3 ⋮ n - 1

=> n - 1 thuộc Ư(-3)

=> n - 1 thuộc {-1; 1; -3; 3}

=> n thuộc {0; 2; -2; 4}

\(C< \frac{2}{3}.\frac{4}{5}......\frac{80}{81}\Rightarrow C.C< \frac{C.2....80}{3.5....81}=\frac{1.2.3....79.80}{2.3.4....81}=\frac{1}{81}=\left(\frac{1}{9}\right)^2mà:C>0\Rightarrow C< \frac{1}{9}\)

Shitbo ơi em có thể giải theo cách cấp 1 được không?

a) \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4}{5}\)

\(\Leftrightarrow2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4}{5}\)

\(\Leftrightarrow2\times\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4}{5}\)

\(\Leftrightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4}{5}\)

\(\Leftrightarrow2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{4}{5}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{4}{5}:2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{5}-\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{4}{10}-\frac{5}{10}=\frac{-1}{10}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{4}{10}-\frac{5}{10}=\frac{1}{-10}\)

\(\Leftrightarrow x+1=-10\)

\(\Leftrightarrow x=-10-1\)

\(\Leftrightarrow x=-11\)

Hông chắc !!! <3

b) Đề khó hiểu vậy, nếu đề là : \(x+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)thì làm như sau nha

\(x+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)

\(\Leftrightarrow x+\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)=1\)

\(\Leftrightarrow x+1=1\)

\(\Leftrightarrow x=1-1\)

\(\Leftrightarrow x=0\)

Rất vui vì giúp đc bạn <3

Cô mk giao thế, bó tay.com. Ko bỏ trị tuyệt đối đi vô lý như thế chứ

\(\left(x+\frac{5}{3}\right)\left(x-\frac{5}{4}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{5}{3}\right)=0\\\left(x-\frac{5}{4}\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{5}{4}\end{cases}}}\)

\(\frac{1}{3}+\frac{1}{2}\div x=-4\)

\(\frac{1}{2}\div x=-4-\frac{1}{3}\)

\(\frac{1}{2}\div x=-\frac{13}{3}\)

\(x=-\frac{3}{26}\)

\(\frac{5}{12}+\left(-\frac{7}{12}\right):x=-\frac{3}{10}\)

\(\Rightarrow-\frac{7}{12}:x=-\frac{43}{60}\)

\(\Rightarrow x=\frac{35}{43}\)

\(\frac{5}{12}+-\frac{7}{12}:x=\frac{-3}{10}\)

\(\Leftrightarrow\)\(-\frac{7}{12}:x=\frac{-3}{10}-\frac{5}{12}\)

\(\Leftrightarrow\)\(-\frac{7}{12}:x=\frac{-18}{60}-\frac{25}{60}\)

\(\Leftrightarrow\)\(-\frac{7}{12}:x=\frac{-43}{60}\)

\(\Leftrightarrow\) \(x=\frac{-7}{12}:\frac{-43}{60}\)

\(\Leftrightarrow\) \(x=\frac{-7}{12}.\frac{60}{-43}\)

\(\Leftrightarrow\) \(x=\frac{35}{43}\)

Vậy \(x=\frac{35}{43}\)