a.

\(x^2-4xy=23\)

\(\Leftrightarrow x\left(x-4y\right)=23\)

Ta co:

\(23=1.23=23.1=\left(-1\right).\left(-23\right)=\left(-23\right).\left(-1\right)\)

TH1:

\(\left\{{}\begin{matrix}x=1\\x-4y=23\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\frac{11}{2}\end{matrix}\right.\)(loai)

TH2:

\(\left\{{}\begin{matrix}x=23\\x-4y=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=23\\y=\frac{11}{2}\end{matrix}\right.\)(loai)

TH3:

\(\left\{{}\begin{matrix}x=-1\\x-4y=-23\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\frac{11}{2}\end{matrix}\right.\)(loai)

TH4:

\(\left\{{}\begin{matrix}x=-23\\x-4y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-23\\y=-\frac{11}{2}\end{matrix}\right.\)(loai)

Vay khong co ngiem nguyen nao thoa man phuong trinh

\(\hept{\begin{cases}3x-y=2m+3\\x+2y=3m+1\end{cases}}\Leftrightarrow\hept{\begin{cases}6x-2y=4m+6\\x+2y=3m+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=m+1\\y=m\end{cases}}\)khi đó: \(^{x^2+y^2=5\Leftrightarrow2m^2+2m+1=5\Leftrightarrow2m^2+2m-4=0\Leftrightarrow\orbr{\begin{cases}m=1\\m=-2\end{cases}}}\)

xy-x-y=2

=> x.(y-1)-y=2

=>x.(y-1)-(y-1)=3

=>(x-1)(y-1)=3

=> x-1 và y-1 thuộc Ư(3)

Ư(3)={-3;-1;1;3}

Ta có bảng

pt : 

\(2x^6-2x^3y+y^2=320\Leftrightarrow x^6+\left(x^6-2x^3y+y^2\right)=320\)

\(\Leftrightarrow x^6+\left(x^3-y\right)^2=320\)

=> \(x^6\le320\Leftrightarrow-2\le x\le2\)

TH1: Nếu \(x=-2\Rightarrow x^6=64\Rightarrow\left(x^3-y\right)^2=320-64=256\Rightarrow\orbr{\begin{cases}x^3-y=-16\\x^3-y=16\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=x^3+16=\left(-2\right)^3+16=8\\y=x^3-16=\left(-2\right)^3-16=-24\end{cases}}\)

TH2: Nếu \(x=2\Rightarrow x^6=64\Rightarrow\left(x^3-y\right)^2=320-64=256\Rightarrow\orbr{\begin{cases}x^3-y=-16\\x^3-y=16\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=x^3+16=2^3+16=24\\y=x^3-16=2^3-16=-8\end{cases}}\)

TH3: Nếu \(\orbr{\begin{cases}x=-1\\x=1\end{cases}}\Rightarrow x^6=1\Rightarrow\left(x^3-y\right)^2=320-1=319\) (vô nghiệm nguyên)

TH4: Nếu \(x=0\Rightarrow x^6=0\Rightarrow\left(x^3-y\right)^2=320\)(vô nghiệm nguyên)

Vậy pt có nghiệm (x,y)=...

(2x-1)⁵=(2x-1)⁷

=>x=0;1;\(\frac{1}{2}\)

(x-2)².(y-3)²=4

phân tích rồi kẻ bảng ra

Xét \(\Delta=m^2+8>0\) nên phương trình luôn có nghiệm

Theo Viete \(x_1+x_2=-m\left(1\right);x_1x_2=-2\)

Mà \(x_1^2=4x_2^2\Leftrightarrow x_1=2x_2\left(h\right)x_1=-2x_2\)

Bạn thay vào ( 1 ) là ra pt bậc nhất 1 ẩn,khi đó dể nè :))