Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}=\dfrac{\left(a-a\right)+\left(b-b\right)+\left(c-c\right)+a+b+c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
Hay \(\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\b+c-a=a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)
Thay vào \(M\) ta có:
\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2c.2a.2b}{abc}=\dfrac{8abc}{abc}=8\)
2) Ta có:
\(3n+2-2n+2+3n-2n=\left(3n-2n+3n-2n\right)+\left(2+2\right)=2n+4⋮̸10\)
Đề sai