\(\left(x+3\right)^{2n+1}=\left(2x-3\right)^{2n+1}\)

\(\Rightarrow x+3=2x-3\)

\(x+3-2x+3=0\)

\(\left(x-2x\right)+\left(3+3\right)=0\)

\(-x+6=0\)

\(-x=0-6\)

\(-x=-6\)

\(\Rightarrow x=6\)

Vậy \(x=6\).

Từ đẳng thức \(7\cdot\left(-28\right)=\left(-49\right)\cdot4\), ta có các tỉ lệ thức sau:

\(\dfrac{7}{4}=\dfrac{-49}{-28}\);\(\dfrac{4}{7}=\dfrac{-28}{-49}\);\(\dfrac{7}{-49}=\dfrac{4}{-28}\);\(\dfrac{-49}{7}=\dfrac{-28}{4}\)

choáng,vào viện đột ngột mất rồi.

\(\left(\dfrac{1}{2}\right)^3\cdot\left[\left(\dfrac{1}{2}\right)^x\right]^x-\dfrac{5}{8}=\dfrac{1}{16}\cdot\left(-9\right)\)

\(\left(\dfrac{1}{2}\right)^3\cdot\left[\left(\dfrac{1}{2}\right)^x\right]^x-\dfrac{5}{8}=\dfrac{-9}{16}\)

\(\left(\dfrac{1}{2}\right)^3\cdot\left[\left(\dfrac{1}{2}\right)^x\right]^x=\dfrac{-9}{16}+\dfrac{5}{8}\)

\(\left(\dfrac{1}{2}\right)^3\cdot\left[\left(\dfrac{1}{2}\right)^x\right]^x=\dfrac{1}{16}\)

\(\left(\dfrac{1}{2}\right)^3\cdot\left[\left(\dfrac{1}{2}\right)^x\right]^x=\left(\dfrac{1}{2}\right)^4\)

\(\left[\left(\dfrac{1}{2}\right)^x\right]^x=\left(\dfrac{1}{2}\right)^4\div\left(\dfrac{1}{2}\right)^3\)

\(\left[\left(\dfrac{1}{2}\right)^x\right]^x=\dfrac{1}{2}\)

\(\left[\left(\dfrac{1}{2}\right)^x\right]^x=\left[\left(\dfrac{1}{2}\right)^1\right]^1\)

\(\Rightarrow x=1\)