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ta có : \(x+3y=xy+3\Leftrightarrow x+3y-xy-3\Leftrightarrow-xy+3y+x-3\)
\(\Leftrightarrow-y\left(x-3\right)+\left(x-3\right)=\left(1-y\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}1-y=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=3\end{matrix}\right.\) vậy \(y=1;x=3\)
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\(\left(2x+y\right)^2=4x^2+4xy+y^2\)
\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)
\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{y^2}{4}\)
\(\left(x-2y\right)^2\left(x+2y\right)^2=\left(x^2-4y^2\right)^2\)
\(=x^4-8x^2y^2+16y^4\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)
\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)
\(\left(2x+y\right)^2=4x^2+4xy+y^2\)
\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)
\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{x^2y}{2}+\frac{x^2y}{2}-\frac{y^2}{4}=x^4-\frac{y^2}{4}\)
\(\left(x-2y\right)^2\left(x+2y\right)^2=x^4-8x^2y^2+16y^4\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-xy^2+xy^2-1=x^2y^4-1\)
\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+3y=xy+3\)
\(\Leftrightarrow x+3y-xy-3=0\)
\(\Leftrightarrow x-xy+3y-3=0\)
\(\Leftrightarrow x\left(1-y\right)-3\left(1-y\right)=0\)
\(\Leftrightarrow\left(1-y\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-y=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1\\x=3\end{matrix}\right.\)
Vậy phương trình trên bằng nhau xảy ra khi
\(x=3\) và \(y=1\)
Hệ số của x2y2 trong khai triển (2x - 3y2)3 là??? Tham khảo mình làm đây này đúng 100%
= -36