Theo đầu bài ta có:
\(\left(2x-7\right)^{24}+\left(7-2x\right)^{24}=2\)
\(\Rightarrow\left(2x-7\right)^{24}+\left[-\left(2x-7\right)\right]^{24}=2\)
\(\Rightarrow\left(2x-7\right)^{24}+\left(2x-7\right)^{24}=2\)
\(\Rightarrow\left(2x-7\right)^{24}\cdot2=2\)
\(\Rightarrow\left(2x-7\right)^{24}=1\)
\(\Rightarrow\orbr{\begin{cases}2x-7=1\\2x-7=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=8\\2x=6\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)

\(\frac{x}{4}=\frac{y}{7}=k\)

\(\Rightarrow\hept{\begin{cases}x=4k\\y=7k\end{cases}}\)

xy = 112

=> 4k . 7k = 112

=> 28 . k² = 112

=> k² = 4 \(\Rightarrow\hept{\begin{cases}k=4\\k=-4\end{cases}}\)

=> \(\hept{\begin{cases}k=4\Rightarrow\hept{\begin{cases}x=16\\y=28\end{cases}}\\k=-4\Rightarrow\hept{\begin{cases}x=-16\\y=-28\end{cases}}\end{cases}}\)

Theo đầu bài ta có:
\(\frac{x}{4}=\frac{y}{7}\)
\(\Rightarrow\left(\frac{x}{4}\right)^2=\frac{x}{4}\cdot\frac{y}{7}=\frac{112}{28}=4\)
\(\Rightarrow\frac{x}{4}=\frac{y}{7}=\sqrt{4}=2\)
\(\Rightarrow x=2\cdot4=8\)
\(\Rightarrow y=2\cdot7=14\)

\(5x\left(2x-\frac{1}{2}\right)+2\left(2x-\frac{1}{2}\right)=0\)

<=>\(\left(2x-\frac{1}{2}\right)\left(5x+2\right)=0\)

<=>\(\orbr{\begin{cases}2x-\frac{1}{2}=0\\5x+2=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=\frac{1}{4}\\x=\frac{-2}{5}\end{cases}}\)

Vậy x=1/4 hoặc -2/5

(x - 3)⁴ =16

=> 2⁴ = 16

=> x- 3 = 2

=> x = 5

4^(x-3) =16

=> 4² = 16

=> x-3 = 2 

=> x = 5

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