ĐK : \(x;y\in Z;y\ge0\)

\(\sqrt{x^2-2x+13}=y\)

\(\Leftrightarrow x^2-2x+13=y^2\)

\(\Leftrightarrow\left(x^2-2x+1\right)+12=y^2\)

\(\Leftrightarrow\left(x-1\right)^2+12=y^2\)

\(\Leftrightarrow\left(x-1\right)^2-y^2=-12\)

\(\Leftrightarrow\left(x-y-1\right)\left(x+y-1\right)=-12\) đến đây lm tiếp

Làm tiếp hộ mk !!!! xem mk làm có đúng ko ><

4:

(x+1)(y-2)=5

=>\(\left(x+1;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;7\right);\left(4;3\right);\left(-2;-3\right);\left(-6;1\right)\right\}\)

a/ ta có: 

\(x\sqrt{2y-1}+y\sqrt{2x-1}=\sqrt{x}.\sqrt{2xy-x}+\sqrt{y}.\sqrt{2xy-y}\)

\(\le\frac{x+2xy-x}{2}+\frac{y+2xy-y}{2}=2xy\)

Dấu = xảy ra khi ...

Khi gì

b/ \(x^4-x^2+2x+2=\left(x+1\right)^2\left(x^2-2x+2\right)\)

\(\Rightarrow x^2-2x+2=y^2\)

Đơn giản rồi ha

\(\Leftrightarrow y^2=x^2+4x+5\left(y\ge0\right)\\ \Leftrightarrow y^2-\left(x+2\right)^2=1\\ \Leftrightarrow\left(y-x-2\right)\left(y+x+2\right)=1\)

Vì \(x,y\in Z\Leftrightarrow\left(y-x-2\right)\left(y+x+2\right)=1\cdot1=\left(-1\right)\left(-1\right)\)

\(\left\{{}\begin{matrix}y-x-2=1\\y+x+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-x=3\\y+x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\left(tm\right)\)

\(\left\{{}\begin{matrix}y-x-2=-1\\y+x+2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-x=1\\y+x=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\left(ktm\right)\)

Vậy \(\left(x;y\right)=\left(-2;1\right)\)

Ai giúp em với ạ

1. Ta có: \(x^2-2xy-x+y+3=0\)

<=> \(x^2-2xy-2.x.\frac{1}{2}+2.y.\frac{1}{2}+\frac{1}{4}+y^2-y^2-\frac{1}{4}+3=0\)

<=> \(\left(x-y-\frac{1}{2}\right)^2-y^2=-\frac{11}{4}\)

<=> \(\left(x-2y-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=-\frac{11}{4}\)

<=> \(\left(2x-4y-1\right)\left(2x-1\right)=-11\)

Th1: \(\hept{\begin{cases}2x-4y-1=11\\2x-1=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-3\end{cases}}\)

Th2: \(\hept{\begin{cases}2x-4y-1=-11\\2x-1=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)

Th3: \(\hept{\begin{cases}2x-4y-1=1\\2x-1=-11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)

Th4: \(\hept{\begin{cases}2x-4y-1=-1\\2x-1=11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=3\end{cases}}\)

Kết luận:...

2. \(y^2+1\ge1>0;2x^2+x+1>0\) với mọi x; y 

=> x + 5 > 0 

=>  \(y^2+1=\frac{x+5}{2x^2+x+1}\ge1\)

<=> \(x+5\ge2x^2+x+1\)

<=> \(x^2\le2\)

Vì x nguyên => x = 0 ; x = 1; x = -1 

Với x = 0 ta có: \(y^2+1=5\Leftrightarrow y=\pm2\)

Với x = 1 ta có: \(y^2+1=\frac{3}{2}\)loại vì y nguyên 

Với x = -1 ta có: \(y^2+1=2\Leftrightarrow y=\pm1\)

Vậy Phương trình có 4 nghiệm:...