a) Để y nguyên thì \(6x-4⋮2x+3\)

\(\Leftrightarrow-13⋮2x+3\)

\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)

\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)

hay \(x\in\left\{-1;-2;5;-8\right\}\)

3/ Ta có:

\(A=\dfrac{1-2x}{x+3}\)

\(A=\dfrac{-2x+1}{x+3}\)

\(A=\dfrac{-2x-6+7}{x+3}\)

\(A=\dfrac{-2\left(x+3\right)+7}{x+3}\)

\(A=-2+\dfrac{7}{x+3}\)

A nguyên khi \(\dfrac{7}{x+3}\) nguyên 

⇒ 7 ⋮ \(x+3\)

\(\Rightarrow x+3\inƯ\left(7\right)\)

\(\Rightarrow x+3\in\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{-2;-4;4;-10\right\}\)

Ta có: \(\frac{x}{2}=\frac{2}{y}+\frac{11}{6}\)

=>\(\frac{3x}{6}-\frac{11}{6}=\frac{2}{y}\)

=>\(\frac{3x-11}{6}=\frac{2}{y}\)

=>y(3x-11)=12

=>(3x-11;y)∈{(1;12);(12;1);(-1;-12);(-12;-1);(2;6);(6;2);(-2;-6);(-6;-2);(3;4);(4;3);(-3;-4);(-4;-3)}

=>(3x;y)∈{(12;12);(23;1);(10;-12);(-1;-1);(13;6);(17;2);(9;-6);(5;-2);(14;4);(15;3);(8;-4);(7;-3)}

mà 3x⋮3

nên (3x;y)∈{(12;12);(9;-6);(15;3)}

=>(x;y)∈{(4;12);(3;-6);(5;3)}

b) Ta quy đồng rồi => x+xy = 4

=> x(y+1) = 4 thì  

Bài 1:

x/-3=9/4

nên x=-9/4*3=-27/4

2x+y=-4

=>y=-4-2x=-4-2*(-27/4)=-4+27/2=27/2-8/2=19/2

Bài 1:

\(\frac{x}{-3}=\frac94\)

=>\(x=-3\cdot\frac94=-\frac{27}{4}\)

2x+y=-4

=>\(y-\frac{27}{2}=-4\)

=>\(y=-4+\frac{27}{2}=\frac{27}{2}-\frac82=\frac{19}{2}\)

Bài 2:

\(\frac{5a+7b}{6a+5b}=\frac{29}{28}\)

=>29(6a+5b)=28(5a+7b)

=>174a+145b=140a+196b

=>34a=51b

=>2a=3b

=>a=3; b=2