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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(x^2-2xy+2y^2-2x+6y+5=\left(x^2-xy+y^2\right)+y^2-2\left(x-y\right)+4y+5\)
\(=\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+\left(y^2+4y+4\right)\)
\(=\left(x-y-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-y=1\\y=-2\end{cases}\Rightarrow\hept{\begin{cases}x=y+1=-1\\y=-2\end{cases}}}\)
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\(x^2+xy-2012x-2013y-2014=0\)
\(\Rightarrow x\left(x+y\right)-2013\left(x+y\right)+\left(x-2013\right)-1=0\)
\(\Rightarrow\left(x-2013\right)\left(x+y+1\right)=1\)
Ta có bảng sau
x-2013 | 1 | -1 | ||
x+y+1 | 1 | -1 | ||
x | 2014 | 2012 | ||
y | -2014 | -2014 |
Ngu quá nên kẻ thừa, thông cảm :)
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+) \(f\left(x\right)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0\)
+) \(f\left(x\right)=p\left(x\right)\left(x-\alpha\right)\)
lược đồ:
\(\alpha\) | \(a_n\) | \(a_{n-1}\) | \(a_{n-2}\) | ... | \(a_1\) | \(a_0\) |
\(\alpha\) | \(b_n=a_n\) | \(b_{n-1}=\alpha b_n+a_{n-1}\) | \(b_{n-2}=\alpha b_{b-1}+a_{n-2}\) | ... | \(b_1=\alpha b_2+a_1\) | \(r=\alpha b_1+a_0\) |
rồi giải đi :v
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2-5xy+2y^2=0\)
\(\Leftrightarrow\left(2x-y\right)\left(x-2y\right)=0\Rightarrow\left[{}\begin{matrix}y=2x\\x=2y\end{matrix}\right.\)
Mà \(y>x>0\Rightarrow y=2x\)
\(\Rightarrow\frac{2012x+2013y}{3x-2y}=\frac{2012x+2013.2x}{3x-2.2x}=-6038\)