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=>x(1+2y)+y+1/2=49,5
=>(y+1/2)(2x+1)=49,5
=>(2x+1)(2y+1)=99
=>\(\left(2x+1;2y+1\right)\in\left\{\left(1;99\right);\left(3;33\right);\left(9;11\right);\left(11;9\right);\left(33;3\right);\left(99;1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;16\right);\left(16;1\right);\left(4;5\right);\left(5;4\right)\right\}\)
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Tu de bai suy ra 2y+2x=xy<=>...<=>y(2-x)= -2x<=>y=2x/(x-2)<=>y=(2x-4+4)/(x-2)<=>y=2+4/(x-2)
vi x la so nguyen Dưỡng nen x-2 la so nguyen duong va la ước cua 4 => x-2 =1 hoặc x-2= 4 => x=3 hoac x=6
Voi x=3 => y= 6
voi x=6=> y=3
vay cac cap so nguyen duong (x;y) can tim la (3;6); (6;3)
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PT <=> \(x^3+x^2+x^2+1=4^y\)
<=> \(\left(x+1\right)\left(x^2+1\right)=4^y\)
Gọi \(\left(x+1;x^2+1\right)=d\)
=> \(\left\{{}\begin{matrix}x+1⋮d\\x^2+1⋮d\end{matrix}\right.\)
=> 2 \(⋮d\)
=> d\(\in\left\{1;2\right\}\)
TH1: d = 1
=> \(\left[{}\begin{matrix}x+1=1< =>x=0\left(l\right)\\x^2+1=1< =>x=0\left(l\right)\end{matrix}\right.\)
TH2: d = 2
=> x + 1 và x2 + 1 là lũy thừa của 2
Mà \(x^2+1\ge x+1\)
=> \(x^2+1⋮x+1\)
<=> 2 \(⋮x+1\)
Mà x nguyên dương => x + 1 > 1
=> x + 1 = 2
<=> x = 1
<=> y = 1
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