Bài 1: Tìm x, y nguyên biết :

a) 4x + 2xy + y = 7

   => 2.x(y-2)+(y-2)=5

    => ( y-2)(2x+1)= 5

    Ta có bảng sau:

Điều kiện: t/m

Vậy:....

phần b và c tương tự

b: =>x(3-y)+2y-6=-2

=>-x(y-3)+2(y-3)=-2

=>(y-3)(x-2)=2

=>\(\left(x-2;y-3\right)\in\left\{\left(1;2\right);\left(2;1\right);\left(-1;-2\right);\left(-2;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;5\right);\left(4;4\right);\left(1;1\right);\left(0;2\right)\right\}\)

c: =>x(3y+2)+y+2/3=-4+2/3=-10/3

=>(y+2/3)(3x+1)=-10/3

=>(3x+1)(3y+2)=-10

=>\(\left(3x+1;3y+2\right)\in\left\{\left(1;-10\right);\left(10;-1\right);\left(-2;5\right);\left(-5;2\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;-4\right);\left(3;-1\right);\left(-1;1\right);\left(-2;0\right)\right\}\)

\(\Leftrightarrow2x^2+\left(y+1\right)^2+3x\left(y+1\right)+1=0\)
Đặt y+1=a
\(\Rightarrow2x^2+a^2+3ax=-1\)
\(\Leftrightarrow\left(2x+a\right)\left(x+a\right)=-1\)
Tự giải tiếp

                                                      Giúp tôi giải toán Nguyễn Nhật Minh

trong chtt lm j có mà m.n bảo có z

3xy+2y=2-x

=>3xy+2y+x=2

=>\(y\left(3x+2\right)+x+\dfrac{2}{3}=2+\dfrac{2}{3}=\dfrac{8}{3}\)

=>\(3y\left(x+\dfrac{2}{3}\right)+\left(x+\dfrac{2}{3}\right)=\dfrac{8}{3}\)

=>\(\left(x+\dfrac{2}{3}\right)\left(3y+1\right)=\dfrac{8}{3}\)

=>\(\left(3x+2\right)\left(3y+1\right)=8\)

=>\(\left(3x+2;3y+1\right)\in\left\{\left(1;8\right);\left(8;1\right);\left(-1;-8\right);\left(-8;-1\right);\left(2;4\right);\left(4;2\right);\left(-2;-4\right);\left(-4;-2\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-\dfrac{1}{3};\dfrac{7}{3}\right);\left(2;0\right);\left(-1;-3\right);\left(-\dfrac{10}{3};-\dfrac{2}{3}\right);\left(0;1\right);\left(\dfrac{2}{3};\dfrac{1}{3}\right);\left(-\dfrac{4}{3};-\dfrac{5}{3}\right);\left(-2;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;0\right);\left(-1;-3\right);\left(0;1\right);\left(-2;-1\right)\right\}\)

\(2x^2+3xy-2y^2=7\)

\(\Leftrightarrow2x^2-xy+4xy-2y^2=7\)

\(\Leftrightarrow x\left(2x-y\right)+2y\left(2x-y\right)\)

\(\Leftrightarrow\left(x+2y\right)\left(2x-y\right)=7\)

Nếu 2x - y = 7 và x + 2y = 1 thì:

\(2\left(2x-y\right)+x+2y=15\)

\(\Leftrightarrow5x=15\)

\(\Leftrightarrow x=3;y=1\)( thỏa mãn )

Nếu 2x - y = 1 và x + 2y = 7 thì:

\(2\left(2x-y\right)+x+2y=9\)

\(\Leftrightarrow5x=9\Leftrightarrow x=\frac{9}{5}\)( loại )

Nếu 2x - y = -7 và x + 2y = -1 thì:

\(2\left(2x-y\right)+x+2y=-15\)

\(\Leftrightarrow5x=-15\)

\(\Leftrightarrow x=-3;y=1\)( thỏa mãn )

Nếu 2x - y = -1 và x + 2y = -7

\(\Leftrightarrow2\left(2x-y\right)+x+2y=-9\)

\(\Leftrightarrow5x=-9\Leftrightarrow x=\frac{-9}{5}\)( loại )