giup vsssssss mn

bn ơi bn lm đc bài này ko giúp mik vs

tìm x;y trong phương trình nghiệm nguyên sau:

a)x^2+y^2-2.(3x-5y)=11                b)x^2+4y^2=21+6x

We have equation \(x+y=xy\)

\(\Rightarrow xy-x-y=0\)

\(\Rightarrow x\left(y-1\right)-\left(y-1\right)=1\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=1=\left(-1\right).\left(-1\right)=1.1\)

So equation has two value \(\left(2;2\right),\left(0;0\right)\)

We have \(p\left(x+y\right)=xy\)

\(\Leftrightarrow xy-px-py=0\)

\(\Leftrightarrow xy-px-py+p^2=p^2\)

\(\Leftrightarrow x\left(y-p\right)-p\left(y-p\right)=p^2\)

\(\Leftrightarrow\left(x-p\right)\left(y-p\right)=p^2\)

But p is prime so \(Ư\left(p^2\right)=\left\{1;p;p^2\right\}\)

\(\Rightarrow\left(x-p\right)\left(y-p\right)=1.p^2=p.p=p^2.1=\left(-p\right).\left(-p\right)\)

\(=\left(-1\right).\left(-p^2\right)=\left(-p^2\right).\left(-1\right)\)

So equation has values \(S=\left(p+1;p^2+p\right);\left(2p;2p\right);\left(p^2+p;p+1\right);\left(0;0\right)\)

\(;\left(p-1;p-p^2\right);\left(p-p^2;p-1\right)\)

We put \(K=\frac{2n^2+3n+3}{2n-1}\)

We have \(2n^2+3n+3=\left(2n^2-n\right)+2\left(2n-1\right)+5\)

\(=n\left(2n-1\right)+2\left(2n-1\right)+5=\left(n+2\right)\left(2n-1\right)+5\)

So \(K=n+2+\frac{5}{2n-1}\)

\(K\inℤ\Leftrightarrow5⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Prints:

So \(x\in\left\{1;0;3;-2\right\}\)then \(\frac{2n^2+3n+3}{2n-1}\in Z\)

\(pt\Leftrightarrow x^2-x+2x-2+2y^2-2xy^2+y-xy=1\\ \Leftrightarrow\left(1-x\right)\left(2y^2+y-x-2\right)=1\)

e tự xét 2 th ra

\(xy-x+2y=3\)

\(\Leftrightarrow xy-x+2y-2=1\)

\(\Leftrightarrow x\left(y-1\right)+2\left(y-1\right)=1\)

\(\Leftrightarrow\left(x+2\right)\left(y-1\right)=1\)

\(\Rightarrow x+2=1\) thì \(y-1=1\) \(\Rightarrow x=-1\) thì \(y=2\)

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

Vậy ....................

Đề bài: \(xy-x+2y=3\)

\(\Leftrightarrow\left(x+2\right)y=x+3\)

\(\Leftrightarrow x\left(y-1\right)+2y=3\)

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

\(\Rightarrow x+2\ne0\)\(,\)\(y=\frac{x+3}{x+2}\)

\(\Rightarrow x=-3\)\(,\)\(y=0\)

\(x=-1\)\(,\)\(y=2\)

\(\Leftrightarrow x^2+2xy+y^2-xy-x^2y^2=0\)

\(\Leftrightarrow\left(x+y\right)^2=xy\left(xy+1\right)\)

VT là 1 số chính phương mà vế phải là tích 2 số tự nhiên liên tiếp

\(\Rightarrow\left[{}\begin{matrix}xy=0\\xy+1=0\end{matrix}\right.\)

+ Với \(xy=0\Rightarrow\left(x+y\right)^2=x^2+y^2=0\Rightarrow x=y=0\)

+ Với \(xy+1=0\Rightarrow xy=-1\Rightarrow\left[{}\begin{matrix}x=1;y=-1\\x=-1;y=1\end{matrix}\right.\)