[Toán 10] Hệ phương trình

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ngocthao1995

[tex]\left\{ \begin{array}{l} (x+y)^2-4(x+y)=12 \\ (x-y)^2 - 2(x-y)=3 \end{array} \right.[/tex]

Đặt [TEX]\left{\begin{x+y=a}\\{x-y=b} [/TEX]

[TEX]PT \Leftrightarrow \left{\begin{a^2-4a=12}\\{b^2-2b=3} [/TEX]

[TEX]\left{\begin{\left[\begin{a=6}\\{a=-2}}\\{\left[\begin{b=3}\\{b=-1}} [/TEX]

[TEX]TH1 \left{\begin{x+y=6}\\{x-y=3} [/TEX]

[TEX]TH2 \left{\begin{x+y=6}\\{x-y=-1} [/TEX]

[TEX]TH3 \left{\begin{x+y=-2}\\{x-y=3} [/TEX]

[TEX]TH4 \left{\begin{x+y=-2}\\{x-y=-1} [/TEX]

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