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TP
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ND
1
BD
13 tháng 8 2023
\(a,xy-x-y=2\\ x\left(y-1\right)-y=2\\ x\left(y-1\right)-y+1=2+1\\ x\left(y-1\right)-\left(y-1\right)=3\\ \left(y-1\right)\left(x-1\right)=3\\ Th1:x-1=-1=>x=0\\ y-1=-3=>y=-2\\ Th2:x-1=-3 =>x=-2\\ y-1=-1=> y=0\\ Th3:x-1=3=> x=4\\ y-1=1=>y=2\\ Th4:x-1=1=>x=2\\ y-1=3=>y=4\)
Vậy......
\(b,2x^2+3xy-2y^2=7\\ 2x^2+\left(4xy-xy\right)-2y^2=7\\ x\left(2x-y\right)+2y\left(2x-y\right)=7\\ \left(2x-y\right)\cdot\left(x+2y\right)=7\)
Nếu 2x-y=1; x+2y = 7
=> 2(2x-y) + x + 2y = 9
=> 4x - 2y + x +2y = 9
=> (4x+x) + (2y-2y) = 9
=> 5x + 0 = 9
=> x = 9/5 (ktm)
Nếu 2x-y=7; x+2y = 1
=> 2(2x-y) + x+ 2y = 15
=> 4x - 2y + x +2y =15
=> (4x +x)+ (2y-2y) =15
=> 5x +0 =15
=> x= 3 (tm)
=> y= -1 (Tm)
Nếu 2x-y=-7; x+2y = -1
=> 2(2x-y) + x+ 2y = -15
=> 4x - 2y + x +2y =-15
=> (4x +x)+ (2y-2y) =-15
=> 5x +0 =-15
=> x= -3 (tm)
=> y= 1 (tm)
Nếu 2x-y=-1 ; x+2y = -7
=> 2(2x-y) + x+ 2y = -9
=> 4x - 2y + x +2y = -9
=> (4x +x)+ (2y-2y) =-9
=> 5x +0 =-9
=> x= -9/5 (ktm)
=> y= -1
Vậy.........
9 tháng 1 2022
Bài 1:
Để E nguyên thì \(x+5⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1;9;-5\right\}\)
ND
2
14 tháng 8 2023
a) \(2x^2-3xy-2y^2=2\)
\(\Rightarrow2x^2+xy-4xy-2y^2=2\)
\(\Rightarrow x\left(2x+y\right)-2y\left(2x+y\right)=2\)
\(\Rightarrow\left(2x+y\right)\left(x-2y\right)=2\)
\(\Rightarrow\left(2x+y\right);\left(x-2y\right)\in\left\{-1;1;-2;2\right\}\)
Ta giải các hệ phương trình sau với x;y nguyên
1) \(\left\{{}\begin{matrix}2x+y=-1\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-2\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}2x+y=1\\x-2y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=2\\x-2y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}2x+y=-2\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-4\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}2x+y=2\\x-2y=1\end{matrix}\right.\) \(\left\{{}\begin{matrix}4x+2y=4\\x-2y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(-1;0\right);\left(1;1\right)\right\}\)
14 tháng 8 2023
b) \(xy-y+x=9\)
\(\Rightarrow y\left(x-1\right)+x-1+1=9\)
\(\Rightarrow\left(x-1\right)\left(y+1\right)=8\)
\(\Rightarrow\left(x-1\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(0;-9\right);\left(2;7\right);\left(-1;-5\right);\left(3;3\right);\left(-3;-3\right);\left(5;1\right);\left(-7;-2\right);\left(9;0\right)\right\}\)
NT
1
e) \(x+y+3xy=1\)
\(\Leftrightarrow3x+3y+9xy=3\)
\(\Leftrightarrow3x+9xy+3y=3\)
\(\Leftrightarrow3x\left(1+3y\right)+1+3y=4\)
\(\Leftrightarrow\left(3y+1\right)\left(3x+1\right)=4\)
\(\Leftrightarrow\left(3x+1\right);\left(3y+1\right)\in\left\{-1;1;-2;2;-4;4\right\}\)
\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-\dfrac{2}{3};-\dfrac{5}{3}\right);\left(0;1\right);\left(-1;-1\right);\left(\dfrac{1}{3};\dfrac{1}{3}\right);\left(-\dfrac{5}{3};-\dfrac{2}{3}\right);\left(1;0\right)\right\}\)
\(\Leftrightarrow\left(x;y\right)\in\left\{\left(0;1\right);\left(-1;-1\right);\left(1;0\right)\right\}\)