\(\left\{{}\begin{matrix}5^{x+2}=25^y\\27^y=81\cdot3^{x+4}\end{matrix}\right.\)

Từ \(27^y=81\cdot3^{x+4}\Rightarrow\left(3^3\right)^y=3^4\cdot3^{x+4}\)

\(\Rightarrow3^{3y}=3^{x+8}\)\(\Rightarrow3y=x+8\left(1\right)\)

Lại có: \(5^{x+2}=25^y\Rightarrow5^{x+2}=\left(5^2\right)^y\)

\(\Rightarrow5^{x+2}=5^{2y}\Rightarrow x+2=2y\left(2\right)\)

Thay \(\left(2\right)\) vào \(\left(1\right)\) ta có:

\(\left(1\right)\Leftrightarrow3y=2y+6\Leftrightarrow y=6\)

\(\Rightarrow x+2=2y=2\cdot6=12\Rightarrow x=10\)

1.2^x.2^y=2^×+y=2⁵

=>x+y=5

2.(3^2x)^y=3^2xy

27⁴=(3³)⁴=3^3.4=3¹²

=>2xy=12

=>x.y=6

\(a,x\left(y-2\right)=8\\ \Rightarrow x;\left(y-2\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

Vậy \(\left(x;y\right)=\left(-8;1\right),\left(-4;0\right),\left(-2;-2\right),\left(-1;-6\right),\left(2;6\right),\left(4;4\right),\left(8;3\right)\)

\(b,\left(x-1\right)\left(y-2\right)=9\\ \Rightarrow\left(x-1\right),\left(y-2\right)\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\)

Vậy \(\left(x;y\right)=\left(-8;1\right),\left(-2;-1\right),\left(0;-7\right),\left(2;11\right),\left(4;5\right),\left(10;3\right)\)

1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\ \left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\ \left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\ \frac{98}{99}-x=\frac{7}{2}\\ x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)

2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)

1.