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\(\left(2x+5\right)\left(y-3\right)=22\\ \Rightarrow\left(2x+5\right);\left(y-3\right)\inƯ\left(22\right)=\left\{1;2;11;22\right\}\\ TH1:2x+5=1\Rightarrow x=-2\left(loại\right);\left(y-3\right)=22\Rightarrow y=25\\ TH2:2x+5=2\Rightarrow x=-\dfrac{3}{2}\left(loại\right);\left(y-3\right)=11\Rightarrow y=14\\ TH3:2x+5=11\Rightarrow x=3;\left(y-3\right)=2\Rightarrow y=5\\ TH4:2x+5=22\Rightarrow x=\dfrac{17}{2}\left(loại\right);\left(y-3\right)=1\Rightarrow y=4\\Vậy:\left(x;y\right)=\left(3;5\right)\)
\(\left(4x-1\right)\left(y-3\right)=18\)
\(\Rightarrow\left(4x-1\right);\left(y-3\right)\in U\left(18\right)=\left\{1;2;3;6;9;18\right\}\left(x,y\inℤ^+\right)\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(\dfrac{1}{2};31\right);\left(\dfrac{3}{4};12\right);\left(1;9\right);\left(\dfrac{7}{4};6\right);\left(\dfrac{5}{2};5\right);\left(\dfrac{19}{4};4\right)\right\}\left(x,y\inℤ^+\right)\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(1;9\right)\right\}\left(x,y\inℤ^+\right)\)
y-2x+3xy=24
3xy+y-2x=24
y(3x+1)-2x=24
y(3x+1)-2x-24=0
y(3x+1)-2x-2/3-70/3=0
y(3x+1)-2(x+1/3)-70/3=0
3y(x+1/3)-2(x+1/3)-70/3=0
(x+1/3)(3y-2)=70/3
3(x+1/3)(3y-2)=70
(3x+1)(3y-2)=70
Tự làm nhé^^
\(\left(2x+5\right)\left(y-3\right)=22\)
\(\Rightarrow\left(2x+5\right);\left(y-3\right)\in\left\{1;2;11;22\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(-2;25\right);\left(-\dfrac{3}{2};14\right);\left(3;5\right);\left(\dfrac{17}{2};4\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(3;5\right)\right\}\left(\left(x;y\inℤ^+\right)\right)\)