Question

Giải hệ phương trình:

\(\left\{{}\begin{matrix}x^3y-x^2y+x^2y-y^3=0\\xy+3y-\sqrt{x+3}-2=0\end{matrix}\right.\)

Answer 1

Bạn coi lại đề, dòng đầu tiên ấy, nhìn là thấy sai rồi

Answer 2

ĐKXĐ: ...

\(x^3-x^2y+xy^2-y^3=0\)

\(\Leftrightarrow x^2\left(x-y\right)+y^2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(x-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+y^2=0\Rightarrow x=y=0\left(ktm\right)\\x-y=0\Rightarrow x=y\end{matrix}\right.\)

Thay xuống pt dưới:

\(x^2+3x-\sqrt{x+3}-2=0\)

\(\Leftrightarrow\left(x^2+3x-4\right)-\left(\sqrt{x+3}-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)-\frac{x-1}{\sqrt{x+3}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4-\frac{1}{\sqrt{x+3}+2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x+4=\frac{1}{\sqrt{x+3}+2}\left(1\right)\end{matrix}\right.\)

Xét (1), do \(x\ge-3\Rightarrow\left\{{}\begin{matrix}VT=x+4\ge1\\VP=\frac{1}{\sqrt{x+3}+2}\le\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow VT>VP\)

\(\Rightarrow\left(1\right)\) vô nghiệm