2x-xy+3y=-5
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VT
2
26 tháng 6 2017
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
25 tháng 1 2018
\(a\text{) }\left|2x-5\right|+\left|3y+1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|3y+1\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
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 \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(-\dfrac{1}{2}\right)y=-2\end{matrix}\right.\)
2 tháng 8 2017
\(\left\{{}\begin{matrix}2xy+x+2y=5\\xy+3x-3y=5\end{matrix}\right.\)
\(\Rightarrow2xy+x+2y=xy+3x-3y\)
\(\Rightarrow2xy+x+2y-xy-3x+3y=0\)
\(\Rightarrow\left(2xy-xy\right)+\left(x-3x\right)+\left(2y+y\right)=0\)
\(\Rightarrow xy-2x+3y=0\)
\(\Rightarrow xy-2x+3y-6=-6\)
\(\Rightarrow x\left(y-2\right)+3\left(y-2\right)=-6\)
\(\Rightarrow\left(x+3\right)\left(y-2\right)=-6\)
Xét ước là xong,mấy câu kia tương tự
T
3 tháng 1
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:
| x + 3 | 1 | 11 | -1 | -11 |
| y - 2 | -11 | -1 | 11 | 1 |
| x | -2 | 8 | -4 | -14 |
| y | -9 | 1 | 13 | 3 |
Vì \(x,y\) nguyên nên ta được các cặp giá trị \(\left(x;y\right)\) là:
\(\left(-2;-9\right);\left(8;1\right);\left(-4;13\right);\left(14;3\right)\)
\(Toru\)
TG
1
BT
2 tháng 12 2017
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)
AD
23 tháng 7 2023
\(2x-5\sqrt{xy}+3y\\ =2x-2\sqrt{xy}-3\sqrt{xy}+3y\\ =2\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\\ =\left(2\sqrt{x}-3\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 7 2023
\(2x-5\sqrt{xy}+3y\)
\(=2x-2\sqrt{xy}-3\sqrt{xy}+3y\)
\(=2\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(2\sqrt{x}-3\sqrt{y}\right)\)
\(2x-\left(xy-3y\right)=-5\)
\(2\left(x-3\right)-y\left(x-3\right)=-5-6\)
\(\left(x-3\right)\left(2-y\right)=-11\)
\(=>\left[{}\begin{matrix}x-3=-11\\2-y=1\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x-3=-1\\2-y=11\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-9\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x-3=1\\2-y=-11\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=13\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x-3=11\\2-y=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=14\\y=3\end{matrix}\right.\)
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