a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

\(A=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

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Có: \(A=16xy+\dfrac{1}{xy}-15xy\)

Áp dụng bdt Co-si, ta có:

\(16xy+\dfrac{1}{xy}\ge2\sqrt{16xy.\dfrac{1}{xy}}=8\)

Có \(x+y\ge2\sqrt{xy}< =>xy\le\dfrac{1}{4}\)

=> A \(\ge8-15.\dfrac{1}{4}=\dfrac{17}{4}\)

Dấu "=" xảy ra <=> x = y= \(\dfrac{1}{2}\)

\(\Leftrightarrow6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+20=\dfrac{5\left(x+y\right)\left(xy+3\right)}{xy}\ge\dfrac{5\left(x+y\right)2\sqrt{3xy}}{xy}=10\sqrt{3}\left(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}\right)\)

Đặt \(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=t\ge2\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}=t^2-2\)

\(\Rightarrow6\left(t^2-2\right)+20\ge10\sqrt{3}t\)

\(\Rightarrow3t^2-5\sqrt{3}t+4\ge0\)

\(\Rightarrow\left(\sqrt{3}t-1\right)\left(\sqrt{3}t-4\right)\ge0\)

Do \(t\ge2\Rightarrow\sqrt{3}t-1>0\)

\(\Rightarrow\sqrt{3}t-4\ge0\Rightarrow t\ge\dfrac{4}{\sqrt{3}}\)

\(\Rightarrow t^2\ge\dfrac{16}{3}\Rightarrow t^2-2\ge\dfrac{10}{3}\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge\dfrac{10}{3}\) (do \(\dfrac{x}{y}+\dfrac{y}{x}=t^2-2\))

Vậy \(A_{min}=\dfrac{10}{3}\) khi \(\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

khó thế