Ta có:

\(M=36^{38}+41^{33}=\left(36^{38}-1\right)+\left(41^{33}+1\right)\)

\(\Rightarrow\left\{\begin{matrix}36^{38}-1=\left(36-1\right)\left(36^{37}+36^{36}+...+1\right)\\41^{33}+1=\left(41+1\right)\left(41^{32}-41^{31}+...+1\right)\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}35\left(36^{37}+36^{36}+...+1\right)=5.7.Q⋮7\\42\left(41^{32}-41^{31}+...+1\right)=6.7.Q⋮7\end{matrix}\right.\)\(\Leftrightarrow M⋮7\)

Vậy \(M=36^{38}+41^{33}⋮7\) (Đpcm)

Ta có:

\(P\left(x\right)=x^2+2mx+m^2\)

\(\Leftrightarrow P\left(1\right)=1+2m+m^2\)

\(Q\left(x\right)=x^2+\left(2m+1\right).x+m^2\)

\(\Leftrightarrow Q\left(-1\right)=1-\left(2m+1\right)+m^2=m^2-2m\)

Mà \(P\left(1\right)=Q\left(-1\right)\)

\(\Leftrightarrow1+2m+m^2=m^2-2m\)

\(\Leftrightarrow2m+2m=-1\)

\(\Leftrightarrow4m=-1\)

\(\Leftrightarrow m=\frac{-1}{4}\)

Vậy \(m=\frac{-1}{4}\)

\(3xy+x-y=1\)

\(\Leftrightarrow3xy+x=y+1\)

\(\Leftrightarrow x\left(3y+1\right)=y+1\)

\(\Rightarrow y+1⋮3y+1\)

\(\Rightarrow3y+3⋮3y+1\)

\(\Rightarrow\left(3y+2\right)+2⋮3y+1\)

\(\Rightarrow2⋮3y+1\)

\(\Rightarrow3y+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Với:

\(3y+1=-2\Rightarrow y=-1\Leftrightarrow x=0\)

\(3y+1=-1\Rightarrow y=\frac{-2}{3}\) (loại vì \(y\notin Z\))

\(3y+1=1\Rightarrow y=0\Leftrightarrow x=1\)

\(3y+1=2\Rightarrow y=\frac{1}{3}\) (loại vì \(y\notin Z\))

Vậy có \(2\) cặp số nguyên \(\left(x;y\right)\) là \(\left(0;-1\right),\left(1;0\right)\)