a,3

b,192

\(\left(x-2\right)^7=\left(x-2\right)^5\)

\(\Leftrightarrow\left(x-2\right)^7-\left(x-2\right)^5=0\)

\(\Leftrightarrow\left(x-2\right)^5.\left[\left(x-2\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^5=0\\\left(x-2\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3;x=1\end{cases}}}\)

Vậy \(x\in\left\{1;-1;2\right\}\)

:V sửa lại hộ mình cái kết luận nhé 

\(x\in\left\{-1;3;2\right\}\)

45² + 40² - 15² + 80.45 = (40² + 2.40.45 + 45²) - 15²

                                      = (40+45)² - 15²

                                      = (40+45-15).(40+45+15)

                                      = 70.100 = 7000

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+.......+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+........+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.......+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\left(đpcm\right)\)