\(...\Rightarrow\left(x-4\right)^5-\left(x-4\right)^4=0\)

\(\Rightarrow\left(x-4\right)^4.\left(x-4-1\right)=0\)

\(\Rightarrow\left(x-4\right)^4.\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(\Rightarrow x\in\left\{4;5\right\}\)

\(...\Rightarrow x=\dfrac{3}{8}-\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{9}{24}-\dfrac{16}{24}-\dfrac{12}{24}=-\dfrac{19}{24}\)

\(...\Rightarrow x=\dfrac{17}{5}:134=\dfrac{17}{5}\dfrac{1}{134}=\dfrac{17}{670}\)

\(\dfrac{17}{5}:x=134\\ \Rightarrow x=\dfrac{17}{5}:134\\ \Rightarrow x=\dfrac{17}{670}\)

X-22.6=15.2

         X=15.2+22.6

         X=37.8

\(...=53,7:2=26,85\)

(32,2+21,5):2

=54.7:2

=27.35

\(\dfrac{5}{4}+1,5-\dfrac{2}{3}=\dfrac{11}{4}-\dfrac{2}{3}=\dfrac{25}{12}\)

\(\dfrac{5}{2}\)

\(C=-|x+\dfrac{5}{3}|\le0\)

\(\Rightarrow Max\left(C\right)=0\left(x=-\dfrac{5}{3}\right)\)

\(D=2-|3-x|\)

Ta có :

\(|3-x|\ge0\)

\(\Rightarrow-|3-x|\le0\)

\(\Rightarrow2-|3-x|\le2\)

\(\Rightarrow Max\left(D\right)=2\left(x=3\right)\)

Anh làm rồi em hi

 \(\dfrac{4n^2-9}{2n+3}=\dfrac{\left(2n+3\right)\left(2n-3\right)}{2n+3}=2n-3\)

Để \(\dfrac{4n^2-9}{2n+3}\) là số nguyên

\(\Rightarrow2n-3\in Z\)

\(\Rightarrow\forall n\in Z\)