\(3x^2-6x+4y^2-4xy+4y+3=0\)
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Ta có:
\(3x^2-6x+4y^2-4xy+4y+3=0\)
\(\Leftrightarrow x^2-4xy+4y^2-2x+4y+1+2x^2-4x+2=0\)
\(\Leftrightarrow\left(x-2y\right)^2-2\left(x-2y\right)+1+2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-2y-1\right)^2+2\left(x-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y-1=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
Vậy ...
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\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y-x\right)\left(2y+x\right)}{\left(x-2y\right)^2}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
Điều kiện: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)
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nếu bài yêu cầu giải phương trình thì thế này ạ
\(3x^2-6x+4y^2-4xy+4y+3=0\)
\(x^2+4y^2+1-4xy+4y-6x+2x^2-4x+2=0\)
\(\left(2y-x+1\right)^2+2\left(x-1\right)^2=0\)
mà \(\left(2y-x+1\right)^2,\left(x-1\right)^2\ge0\)
\(\int^{x-1=0}_{2y-x+1=0}\Leftrightarrow\int^{x=1}_{y=0}\)
ai cho mình thêm 4 li-ke cho lên 155 với