[Toán 10] Hệ phương trình

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nguyenbahiep1

[laTEX]\begin{cases} \sqrt{x}+\sqrt{2-y}= \sqrt{2} \\ \sqrt{y}+ \sqrt{2-x} = \sqrt{2} \end{cases}[/laTEX]

lấy hệ (1) - (2)

[laTEX]\sqrt{x} - \sqrt{y} + \sqrt{2-y} - \sqrt{2-x} = 0 \\ \\ \frac{x-y}{\sqrt{x} + \sqrt{y}} + \frac{x-y}{\sqrt{2-y} + \sqrt{2-x}} = 0 \\ \\ (x-y).(\frac{1}{\sqrt{x} + \sqrt{y}} + \frac{1}{\sqrt{2-y} + \sqrt{2-x}}) = 0 \\ \\ x = y \Rightarrow \sqrt{x}+\sqrt{2-x}= \sqrt{2} \\ \\ x = 2 , x = 0 [/laTEX]
 
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