x^2+2x^2y^2+2y^2-(x^2y^2+2x^2)-2=0 tim x va y
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mik ko biết
mong bn thông cảm
nha ................
a) x2+2y2+2xy-2y+1=0
\(\Leftrightarrow\)(x2+2xy+y2)+(y2-2y+1)=0
\(\Leftrightarrow\)(x+y)2+(y-1)2=0
\(\Leftrightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
Vậy x=-1, y=1
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\(a,4x^2+9y^2+4x-24y+17=0\)
\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)
\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)
\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)
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Ta có: \(\left(ax+by\right)^2=\left(a^2+b^2\right)\left(x^2+y^2\right)\)
\(\Leftrightarrow a^2x^2+2abxy+b^2y^2=a^2x^2+a^2y^2+x^2b^2+b^2y^2\)
\(\Leftrightarrow2abxy=a^2y^2+x^2b^2\)
\(\Leftrightarrow\left(ay-xb\right)^2=0\)
\(\Leftrightarrow ay=xb\)
hay \(\dfrac{a}{x}=\dfrac{b}{y}\)
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