Ta có : xy - 2x + 3y - 5 = 0 

<=> x(y - 2) + 3y - 6 + 1 = 0

<=> x(y - 2) + 3(y - 2) + 1 = 0

=> (y - 2) (x + 3) = -1

Suy ra :  (y - 2) (x + 3) thuộc Ư(-1) = {-1;1}

Th1 : nếu y - 2 = -1 thì x + 3 = -1 => y = 1 ; x = -4

Th2 : nếu y - 2 = 1 thì x + 3 = 1 => y = 3 , x = -2 

what the hell???

avatar mèo đen

$\begin{array}{c}xy-2x-3y=5\\ \Rightarrow xy-2x-3y+6=5+6\\ \Rightarrow x\left(y-2\right)-3\left(y-2\right)=11\\ \Rightarrow \left(x-3\right)\left(y-2\right)=11\end{array}$

Vì $11=1.11=\left(-1\right).\left(-11\right)$ nên ta có bảng sau:

Vậy $\left(x;y\right)=\left\{\left(-8;1\right);\left(2;-9\right);\left(4;13\right);\left(14;3\right)\right\}$.

\(a\text{) }\left|2x-5\right|+\left|3y+1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|3y+1\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\left|3x-4\right|+\left|3y-5\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|3y-5\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y-5=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=4\\3y=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{5}{3}\end{matrix}\right.\)
c) \(\left|2x-5\right|+\left|xy-3y+2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|xy-3y+2\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\xy-3y+2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\xy-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\xy-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\dfrac{5}{2}y-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(\dfrac{5}{2}-3\right)y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(-\dfrac{1}{2}\right)y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(-\dfrac{1}{2}\right)y=-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2xy+x+2y=5\\xy+3x-3y=5\end{matrix}\right.\)

\(\Rightarrow2xy+x+2y=xy+3x-3y\)

\(\Rightarrow2xy+x+2y-xy-3x+3y=0\)

\(\Rightarrow\left(2xy-xy\right)+\left(x-3x\right)+\left(2y+y\right)=0\)

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

\(\Rightarrow xy-2x+3y-6=-6\)

\(\Rightarrow x\left(y-2\right)+3\left(y-2\right)=-6\)

\(\Rightarrow\left(x+3\right)\left(y-2\right)=-6\)

Xét ước là xong,mấy câu kia tương tự

bài này của bn giống mk DDT Miner Ter

Ta có: \(xy-2x+3y=-5\)

\(\Rightarrow\left(xy-2x\right)+3y-6=-5-6\)

\(\Rightarrow x\left(y-2\right)+3\left(y-2\right)=-11\)

\(\Rightarrow\left(x+3\right)\left(y-2\right)=-11\)

Vì \(x,y\) nguyên nên \(x+3;y-2\) có giá trị nguyên

\(\Rightarrow x+3;y-2\) là các ước của \(-11\)

Ta có bảng sau:

Vì \(x,y\) nguyên nên ta được các cặp giá trị \(\left(x;y\right)\) là:

\(\left(-2;-9\right);\left(8;1\right);\left(-4;13\right);\left(14;3\right)\)

\(Toru\)

Điều kiện \(x\ne\pm3;y\ne-2\):

 \(P=\frac{2x+3y}{xy+2x-3y-6}-\frac{6-xy}{xy+2x+3y+6}-\frac{x^2+9}{x^2-9}.\)

=> \(P=\frac{2x+3y}{\left(y+2\right)\left(x-3\right)}-\frac{6-xy}{\left(y+2\right)\left(x+3\right)}-\frac{x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{\left(2x+3y\right)\left(x+3\right)-\left(6-xy\right)\left(x-3\right)-\left(x^2+9\right)\left(y+2\right)}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{2x^2+3xy+6x+9y-6x+x^2y+18-3xy-x^2y-9y-2x^2-18}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{0}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}=0\)

=> P=0 (với mọi x khác 3, -3 và y khác -2)

\(2x-5\sqrt{xy}+3y\\ =2x-2\sqrt{xy}-3\sqrt{xy}+3y\\ =2\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\\ =\left(2\sqrt{x}-3\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

\(2x-5\sqrt{xy}+3y\)

\(=2x-2\sqrt{xy}-3\sqrt{xy}+3y\)

\(=2\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(2\sqrt{x}-3\sqrt{y}\right)\)