Cách khác: Ta có \(x^2y+2xy+y=32x\)

\(\Leftrightarrow y\left(x+1\right)^2=32x\).

Từ đó \(32x⋮\left(x+1\right)^2\).

Mà \(\left(x,\left(x+1\right)^2\right)=1\) nên \(32⋮\left(x+1\right)^2\Leftrightarrow\left(x+1\right)^2\in\left\{1;4;16\right\}\).

+) Với \(\left(x+1\right)^2=1\Rightarrow x=0\) (loại)

+) Với \(\left(x+1\right)^2=4\Rightarrow x=1;y=8\)

+) Với \(\left(x+1\right)^2=16\Rightarrow x=3;y=6\).

Vậy...

\(\Leftrightarrow y\left(x^2+2x+1\right)-32x-32=-32\)

\(\Leftrightarrow y\left(x+1\right)^2-32\left(x+1\right)=-32\)

\(\Leftrightarrow\left(x+1\right)\left(xy+y-32\right)=-32\)

Do \(x+1\ge2\) nên chỉ có các trường hợp sau:

TH1: \(\left\{{}\begin{matrix}x+1=2\\xy+y-32=-16\end{matrix}\right.\) 

TH2: \(\left\{{}\begin{matrix}x+1=4\\xy+y-32=-8\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}x+1=8\\xy+y-32=-4\end{matrix}\right.\)

TH4: \(\left\{{}\begin{matrix}x+1=16\\xy+y-32=-2\end{matrix}\right.\)

TH5: \(\left\{{}\begin{matrix}x+1=32\\xy+y-32=-1\end{matrix}\right.\)

Bạn tự giải

        \(x^2y+2xy+y=32x\)

\(\Leftrightarrow y\left(x^2+2x+1\right)=32\left(x+1\right)-32\)

\(\Leftrightarrow y\left(x+1\right)^2=32\left(x+1\right)-32\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(32-xy-y\right)=32\)

Vì x, y nguyên dương nên:

...( tự làm nhé!)

32 nha

\(x^3-32x=-y\left(2x+1\right)\Rightarrow-y=\dfrac{x^3-32x}{2x+1}\)

\(\Leftrightarrow-8y=\dfrac{8x^3-256x}{2x+1}=4x^2-2x-127+\dfrac{127}{2x+1}\)

\(\Rightarrow2x+1=Ư\left(127\right)=\left\{-127;-1;1;127\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=-127\left(loại\right)\\2x+1=-1\left(loại\right)\\2x+1=1\left(loại\right)\\2x+1=127\end{matrix}\right.\) \(\Rightarrow x=63\Rightarrow y=-1953< 0\) (loại)

Pt đã cho không có nghiệm nguyên dương

\(2x^2+2y^2+z^2+2xy+2yz+2xz+32x+34y+545=0\)

\(\Leftrightarrow\left(x^2+2.x.16^2+16^2\right)+\left(y^2+2.y.17+17^2\right)+\left(x^2+y^2+z^2+2xy+2yz+2zx\right)=0\)\(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2=0\)

Ta có: \(\left\{{}\begin{matrix}\left(x+16\right)^2\ge0\forall z\\\left(y+17\right)^2\ge0\forall y\\\left(x+y+z\right)^2\ge0\forall x;y;z\end{matrix}\right.\)\(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2\ge0\forall x;y;z\)

Mà \(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+16\right)^2=0\\\left(y+17\right)^2=0\\\left(x+y+z\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+16=0\\y+17=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=-17\\x+y+z=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=-17\\z-33=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=17\\z=33\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=-16\\y=17\\z=33\end{matrix}\right.\)

\(y\left(x+1\right)^2=32x\)

Nhận thấy \(x=-1\) không phải nghiệm \(\Rightarrow y=\frac{32x}{\left(x+1\right)^2}\)

Ta có \(\left(x+1\right)^2=x\left(x+2\right)+1\Rightarrow\left(x+1\right)^2\) và x nguyên tố cùng nhau

\(\Rightarrow32⋮\left(x+1\right)^2\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=4\\\left(x+1\right)^2=16\end{matrix}\right.\)

\(\Rightarrow x=...\Rightarrow y=...\)

a) Sửa đề: \(3x\left(2xy-x^2\right)+5\left(x-2y\right)^2\)

Ta có: \(3x\left(2xy-x^2\right)+5\left(x-2y\right)^2\)

\(=3x^2\left(2y-x\right)+5\left(2y-x\right)^2\)

\(=\left(2y-x\right)\left[3x^2+5\left(2y-x\right)\right]\)

\(=\left(2y-x\right)\left(3x^2+10y-5x\right)\)

c)Ta có: \(x\left(y-2015\right)+2016\left(2015-y\right)\)

\(=x\left(y-2015\right)-2016\left(y-2015\right)\)

\(=\left(y-2015\right)\left(x-2016\right)\)

a/ (x^2-4x+4)+(y^2+2y+1)=0

<=> (x-2x)^2+(y+1)^2 = 0 Vậy x=2 và y = -1

b/ (x^2+2xy+y^2) + ( y^2-2y+1) = 0 

<=> (x+y)^2 + (y-1)^2 = 0 Vậy x=y=1 

a) { x^2 - 4x +4 } +{y^2+2x+1}=0

<=>{ x - 2x}^2+{y+1}^2=0 Vậy x =2 vầy =-1

b) { x^2 +2xy +y^2} +{y^2 - 2y +1=0}

<=> {x+y}^2+{ y - 1 }^2 =0 Vậy x=y=1.

NHA BẠN!