Bài 1:Giải:

Nếu \(n\) lẻ thì \(2n\equiv-1\) (\(mod\) \(3\))

Từ \(PT\Rightarrow z^2\equiv-1\) ( \(mod\) \(3\)) (loại)

Nếu \(n\) chẵn thì \(n=2m\left(m\in N\right)\)

\(PT\) trở thành:

\(z^2-2^{2m}=153\) Hay \(\left(z-2m\right)\left(z+2m\right)=153\)

\(\Rightarrow z+2m\) và \(z-2m\inƯ\left(153\right)\)

\(\Leftrightarrow\) Ta tìm được: \(\left\{{}\begin{matrix}m=2\\z=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}n=4\\z=13\end{matrix}\right.\)

Vậy \(\left(n;z\right)=\left(4;13\right)\)

Bài 2:

b) Theo đề bài ta có:

\(35\left(x+y\right)=210\left(x-y\right)=12x.y\)

Chia các tích trên cho \(BCNN\left(35;210;12\right)=420\) ta được:

\(\dfrac{35\left(x+y\right)}{420}=\dfrac{210\left(x-y\right)}{420}=\dfrac{12xy}{420}\)

Hay \(\dfrac{x+y}{12}=\dfrac{x-y}{2}=\dfrac{xy}{35}\left(1\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+y}{12}=\dfrac{x-y}{2}=\dfrac{\left(x+y\right)+\left(x-y\right)}{12+2}=\dfrac{\left(x+y\right)-\left(x-y\right)}{12-2}\)

\(\Leftrightarrow\dfrac{x+y}{12}=\dfrac{x-y}{2}=\dfrac{x}{7}=\dfrac{y}{5}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)

\(\Leftrightarrow\dfrac{xy}{35}=\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{xy}{7y}=\dfrac{xy}{5x}\)

Mà \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}7y=35\\5x=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)

Vậy hai số nguyên dương \(x;y\) là \(7;5\)

2) ĐK: x;y ∈ Z

pt ⇔ \(\left(x-y\right)^2+\left(y-1\right)\left(y-3\right)=0\)

=> I) a) x-y=0 => x=y

b) y-1=0 => y=1 => x=y=1(nhận)

II) a) x-y=0 => x=y

b) y-3=0 => y=3 => x=y=3(nhận)

\(A=n^{2015}+n^{1015}+1\)

\(\Rightarrow A=n^2\left[\left(n^3\right)^{671}-1\right]+n\left[\left(n^3\right)^{338}-1\right]+n^2+n+1\)

\(\Rightarrow A=\left(n^2+n+1\right)\left(A.n^2.\left(n-1\right)+B.n\left(n-1\right)+1\right)\)

Với n = 1 (t/m). Với n > 1 \(\Rightarrow A\) là hợp số .

\(1+3^{x+1}+2.3^{3x}=y^3\)

Đặt : \(3^x=t\)\(\Rightarrow2t^3+3t+1=y^3\)

Ta Thấy : \(y\equiv1\left(mod3\right)\)\(\Rightarrow y=3k+1\)

\(\Rightarrow2t\left(t^2+3\right)=27k^3+27k^2+9k\)

Đặt t = 3i (i \(\ge1\))

\(\Rightarrow2i\left(3i^2+1\right)=3k^3+3k^2+k=k\left(3k^2+3k+1\right)\)

Ta Thấy : \(\left(i;3i^2+1\right)=1\) và \(3k^2+3k+1\) luôn lẻ .

\(\Rightarrow\) Các trường hợp :

TH1:

\(\left\{{}\begin{matrix}2i=k\\3i^2+1=3k^2+3k+1\end{matrix}\right.\)\(\Rightarrow9k^2+12k=0\Rightarrow k=0\)(loại)

TH2:

\(\left\{{}\begin{matrix}3k^2+3k+1=2i\left(3i^2+1\right)\\k=1\end{matrix}\right.\)

\(\Rightarrow y=4;x=1\)

TH3:

+)

\(k\left(3k^2+3k+1\right)|i\Rightarrow i=rk\left(3k^2+3k+1\right)\)\(\left(r\ge1\right)\)

\(\Rightarrow2r\left(3i^2+1\right)=1\Rightarrow3i^2+1\le\frac{1}{2}\)(loại)

+)

\(k\left(3k^2+3k+1\right)|\left(3i^2+1\right)\Rightarrow\left(3i^2+1\right)=rk\left(3k^2+3k+1\right)\)

\(\Rightarrow2ir=1\Rightarrow i\le\frac{1}{2}\)(loại)

+)

\(k\left(3k^2+3k+1\right)|2i\Rightarrow2i=rk\left(3k^2+3k+1\right)\)

\(\Rightarrow i\le0\)(loại)

+)

\(k\left(3k^2+3k+1\right)|2\left(3i^2+1\right)\Rightarrow2\left(3i^2+1\right)=rk\left(3k^2+3k+1\right)\)

\(\Rightarrow i\le1\) \(\Rightarrow i=1\)\(\Rightarrow t=3\Rightarrow x=1;y=4\)

Vậy \(x=1;y=4\) thỏa mãn .

#Kaito#

\(3.3^{n-1}.\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)

=> \(3^n.\left(2.3.3^{n+2}\right)-2.3^n\left(3^{n+3}-1\right)=405\)

=> \(3^n\left(2.3^{n+3}\right)-2.3^n\left(3^{n+3}-1\right)=405\)

=> ..........