Question

[Ăâêưưgcg]

Mn ơi giúp mk làm câu ê) nh

Mk cảm ơn nhìu!!!!

Answer 1

[Thắm Nguyễn Thị]

https://hoc24.vn/hoi-dap/question/904876.html

Answer 2

[Ngọc Hưng]

\(\left\{{}\begin{matrix}3\left(x+2\right)^2-7\left(x+y-3\right)=43\\7\left(x+2\right)^2+5\left(x+y-3\right)=15\end{matrix}\right.\) (I)

Đặt \(\left(x+2\right)^2=a;x+y-3=b\)

(I) <=> \(\left\{{}\begin{matrix}3a-7b=43\\7a+5b=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=-4\end{matrix}\right.\)

Với \(a=5\Rightarrow\left(x+2\right)^2=5\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{5}\\x+2=-\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}-2\\x=-\sqrt{5}-2\end{matrix}\right.\)

Với \(b=-4\Rightarrow x+y-3=-4\Leftrightarrow y=-1-x=\left[{}\begin{matrix}-1-\sqrt{5}+2=-\sqrt{5}+1\\-1+\sqrt{5}+2=\sqrt{5}+1\end{matrix}\right.\)Vậy hệ phương trình có nghiệm (x;y) = \(\left(\sqrt{5}-2;-\sqrt{5}+1\right);\left(-\sqrt{5}-2;\sqrt{5}+1\right)\)