Bài 1:
a: TH1: p=1
\(2p^2=2\cdot1^2=2\)
=>Nhận
TH2: p>1
=>\(p^2>1\)
=>\(2p^2\) là hợp số
=>Loại
Vậy: p=1
b: TH1: p=3
p+4=3+4=7; p+8=3+8=11
=>Nhận
TH2: p=3k+1
\(p+8=3k+1+8=3k+9=3\left(k+3\right)⋮3\)
=>Loại
TH3: p=3k+2
\(p+4=3k+2+4=3k+6=3\left(k+2\right)⋮3\)
=>Loại
Vậy: p=3
Bài 2:
a: 6xy-2x+9y=68
=>6xy+9y-2x-3=68
=>3y(2x+3)-(2x+3)=68
=>(2x+3)(3y-1)=68
mà 2x+3 lẻ
nên \(\left(2x+3;3y-1\right)\in\){(1;68);(-1;-68);(17;4);(-17;-4)}
=>(x;y)\(\in\left\{\left(-1;23\right);\left(-2;-\dfrac{67}{3}\right);\left(7;\dfrac{5}{3}\right);\left(-10;-1\right)\right\}\)
mà x;y nguyên
nên \(\left(x;y\right)\in\left\{\left(-1;24\right);\left(-10;-1\right)\right\}\)
b: 3xy-2x+5y=29
=>\(x\left(3y-2\right)+5y-\dfrac{10}{3}=29-\dfrac{10}{3}=\dfrac{77}{3}\)
=>\(3x\left(y-\dfrac{2}{3}\right)+5\left(y-\dfrac{2}{3}\right)=\dfrac{77}{3}\)
=>3x(3y-2)+5(3y-2)=77
=>(3x+5)(3y-2)=77
=>\(\left(3x+5;3y-2\right)\in\left\{\left(1;77\right);\left(77;1\right);\left(-1;-77\right);\left(-77;-1\right);\left(7;11\right);\left(11;7\right);\left(-11;-7\right);\left(-7;-11\right)\right\}\)
=>\(\left(x;y\right)\in\){(-4/3;79/3);(24;1);(-2;-25);(-82/3;1/3);(2/3;13/3);(2;3);(-16/3;-5/3);(-4;-3)}
mà x,y nguyên
nên (x;y)\(\in\){(24;1);(-2;-25);(2;3);(-4;-3)}
c: 3x(y+1)+y+1=7
=>(y+1)(3x+1)=7
mà 3x+1 chia 3 dư 1
nên \(\left(3x+1;y+1\right)\in\left\{\left(1;7\right);\left(7;1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;6\right);\left(2;0\right)\right\}\)
