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a) x,y nguyên => x+4; y-8 nguyên
=> x+4; y-8\(\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
ta có bảng
| x+4 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| x | -10 | -7 | -6 | -5 | -3 | -2 | -1 | 2 |
| y-8 | -1 | -2 | -3 | -6 | 6 | 3 | 2 | 1 |
| y | 7 | 6 | 5 | 2 | 14 | 11 | 10 | 9 |
Vậy (x;y)={(-10;7);(-7;6);(-6;5);(-5;2);(-3;14);(-2;11);(-1;10);(2;9)}
b) 2x+xy+3y+6=10
<=> x(2+y)+3(y+2)=10
<=> (y+2)(x+3)=10
x,y nguyên => y+2; x+3 nguyên
=> y+2; x+3\(\in\)Ư(10)={-10;-5;-2;-1;1;2;5;10}
ta có bảng
| x+3 | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
| x | -13 | -8 | -5 | -4 | -2 | -1 | 2 | 7 |
| y+2 | -1 | -2 | -5 | -10 | 10 | 5 | 2 | 1 |
| y | -3 | -4 | -7 | -12 | 8 | 3 | 0 | -1 |
a: =>3y=6x-1
=>y=2x-1/3
Vậy: (a)//(e)
b: y=-0,5x-4
c: y=1/2x+3
d: =>2y=6-x
=>2y=(6-x)/2=-0,5x+3
f: =>y=0,5x+1=1/2x+1
Vậy: (c)//(f), (d)//(b)
\(\Leftrightarrow\left(2x^2-3\right)y=x^2+1\)
\(\Leftrightarrow y=\dfrac{x^2+1}{2x^2-3}\)
\(y\in Z\Rightarrow2y\in Z\Rightarrow\dfrac{2x^2+2}{2x^2-3}\in Z\Rightarrow1+\dfrac{5}{2x^2-3}\in Z\)
\(\Rightarrow2x^2-3=Ư\left(5\right)=\left\{-1;1;5\right\}\)
\(\Rightarrow x^2=\left\{1;2;4\right\}\Rightarrow x=\left\{1;2\right\}\)
- Với \(x=1\Rightarrow y=-2< 0\left(loại\right)\)
- Với \(x=2\Rightarrow y=1\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
\(2x^2-8x=13-3y^2\)
\(\Leftrightarrow2x^2-8x+8=21-3y^2\)
\(\Leftrightarrow2\left(x-4\right)^2=21-3y^2\) (1)
Do \(2\left(x-4\right)^2\ge0;\forall x\Rightarrow21-3y^2\ge0\)
\(\Rightarrow y^2\le7\Rightarrow y^2=\left\{0;1;4\right\}\)
Mặt khác vế trái của (1) là chẵn, 21 là số lẻ \(\Rightarrow3y^2\) lẻ
\(\Rightarrow y^2\) lẻ \(\Rightarrow y^2=1\Rightarrow y=\pm1\)
Thế vào (1) \(\Rightarrow2\left(x-4\right)^2=18\Rightarrow\left(x-4\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(7;1\right);\left(7;-1\right);\left(1;1\right);\left(1;-1\right)\)