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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\(3xy+x+15y-44=0\)

\(3y\left(x+5\right)+\left(x+5\right)-49=0\)

\(\left(x+5\right)\left(3y+1\right)=49\)

Vì x;y là số nguyên \(\Rightarrow\hept{\begin{cases}x+5\in Z\\3y+1\in Z\end{cases}}\)

Có \(\left(x+5\right)\left(3y+1\right)=49\)

\(\Rightarrow\left(x+5\right)\left(3y+1\right)\in\text{Ư}\left(49\right)=\left\{\pm1;\pm7;\pm49\right\}\)

b tự lập bảng nhé~

pt <=> 9x^2+3y^2+12xy+12x+6y+15 = 0

<=> [(9x^2+12xy+4y^2)+2.(3x+2y).2+4] - (y^2+2y+1) + 12 = 0

<=> [(3x+2y)^2+2.(3x+2y).2+4] -(y+1)^2 = -12

<=> (3x+2y+2)^2 - (y+1)^2 = -12

<=> (3x+2y+2+y+1).(3x+2y+2-y-1) = -12

<=> (3x+3y+3).(3x+y+1) = -12

<=> (x+y+1).(3x+y+1) = -4

Đến đó bạn dùng quan hệ ước bội cho các số nguyên mà giải nha !

Tk mk nha

\(\left(x+y\right)\left(4x^2-4xy+y^2\right)=7\)

mik ngại vít,,,bạn tự lm nốt nha

4x³ + y³ - 3xy² - 7 = 0

4x³ - 4x²y + 4x²y + xy² - 4xy² + y³ = 7

(4x³ - 4x²y + xy²) + (4x²y - 4xy² + y³) = 7

x(4x² - 4xy + y²) + y(4x² - 4xy + y²) = 7

(x + y)(4x2 - 4xy + y²) = 7

(x + y).(2x - y)² = 7

=> .....

\(\Leftrightarrow y^2=x^2+4x+5\left(y\ge0\right)\\ \Leftrightarrow y^2-\left(x+2\right)^2=1\\ \Leftrightarrow\left(y-x-2\right)\left(y+x+2\right)=1\)

Vì \(x,y\in Z\Leftrightarrow\left(y-x-2\right)\left(y+x+2\right)=1\cdot1=\left(-1\right)\left(-1\right)\)

\(\left\{{}\begin{matrix}y-x-2=1\\y+x+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-x=3\\y+x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\left(tm\right)\)

\(\left\{{}\begin{matrix}y-x-2=-1\\y+x+2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-x=1\\y+x=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\left(ktm\right)\)

Vậy \(\left(x;y\right)=\left(-2;1\right)\)

\(\Leftrightarrow2x^2y+y=4x^2+5\)

\(\Leftrightarrow y\left(2x^2+1\right)=4x^2+5\)

\(\Leftrightarrow y=\dfrac{4x^2+5}{2x^2+1}=2+\dfrac{3}{2x^2+1}\)

y nguyên \(\Rightarrow\dfrac{3}{2x^2+1}\) nguyên \(\Rightarrow2x^2+1=Ư\left(3\right)\)

Mà \(2x^2+1\ge1\Rightarrow\left[{}\begin{matrix}2x^2+1=1\\2x^2+1=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=5\\x=1\Rightarrow y=3\\x=-1\Rightarrow y=3\end{matrix}\right.\)

5x²+2y+y²-4x-40=0

△=(-4)²-4.5.(2y+y²-40)

△=16-40y-20y²+800

△=-(784+40y+20y²)

△=-(32y+8y+16y²+4y²+16+4+764)

△=-[(4y+4)²+(2y+2)²+764]<0

=>PHƯƠNG TRÌNH VÔ NGHIỆM.