3xy+3y+2x=5
=>3y(x+1)+2x+2=7
=>(x+1)(3y+2)=7
=>\(\left(x+1\right)\left(3y+2\right)=1\cdot7=7\cdot1=\left(-1\right)\cdot\left(-7\right)=\left(-7\right)\cdot\left(-1\right)\)
=>\(\left(x+1;3y+2\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;\dfrac{5}{3}\right);\left(6;-\dfrac{1}{3}\right);\left(-2;-3\right);\left(-8;-1\right)\right\}\)
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