a, Áp dụng tcdtsbn:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\\ \Rightarrow M=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)
b, \(3^{n+2}-2^{n+2}+3^n-2^n=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=3^n\cdot10-2^n\cdot5⋮5\)
\(3^{n+2}-2^{n+2}+3^n-2^n⋮\left(3-2\right)+\left(3-2\right)=2\)
Mà \(\left(2,5\right)=1\) nên \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
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