Đáp án + Giải thích các bước giải:
`x^2+2xy+y^2-6y+10=0`
`<=>(x^2+2xy+y^2)+(y^2-6y+9)+1=0`
`<=>(x+y)^2+(y+3)^2+1=0`
Vì $\begin{cases}(x+y)^2\ge0\forall x,y\\(y+3)^2\ge0\forall x\end{cases}$
`<=>(x+y)^2+(y+3)^2>=0forallx,y`
`<=>(x+y)^2+(y+3)^2+1>=1forallx,y`
`<=>(x+y)^2+(y+3)^2+1>=1>0`
`<=>` Phương trình vô nghiệm.
Đáp án:
Giải thích các bước giải:
$x^{2}+2xy+y^{2}-6y+10=0$
⇔$(x^2+2xy+y^2)+(y^2-6y+9)+1-y^2=0$
⇔$(x+y)^2+(y-3)^2+1-y^2=0$
⇔$(x+y)^2+(y-3)^2+1=y^2$
⇔$(x+y)^2+(y-3)^2≥0$
⇔$(x+y)^2+(y-3)^2+1≥1$
⇔$y^2≥1$
⇔\(\left[ \begin{array}{l}1≤y\\y≤-1\end{array} \right.\)
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