Gõ đề có sai không ạ?

\(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^4\left(1-2x^2\right)=y^4\\1+\sqrt{1+\left(x-y\right)^2}=x^3\left(x^3-x+2y^2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{4-\left(1-x^2y\right)^2}=2x^6-x^4+y^4\\-\sqrt{1+\left(x-y\right)^2}=1-x^6+x^4-2x^3y^2\end{matrix}\right.\)

Cộng theo vế HPT²

\(\sqrt{4-\left(1-x^2y\right)^2}-\sqrt{1+\left(x-y\right)^2}=\left(x^3-y^2\right)^2+1\)

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

Có:

\(\left\{{}\begin{matrix}\sqrt{4-\left(1-x^2y\right)^2}\le2\\\sqrt{1+\left(x-y\right)^2}+\left(x^2-y^2\right)^2+1\ge2\end{matrix}\right.\)

\(\Rightarrow\) (1) xảy ra \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\sqrt{4-\left(1-x^2y\right)^2}=2\\\sqrt{1+\left(x-y\right)^2}=1\\\left(x^3-y^2\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=1\)

b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+y}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) thì pt đầu trở thành:

\(\dfrac{a^2-b^2}{2}-4b^2+3b=a\Leftrightarrow a^2-9b^2+6b=2a\)

\(\Leftrightarrow\left(a-3b\right)\left(a+3b\right)-2\left(a-3b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a+3b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=3b\\a=2-3b\end{matrix}\right.\) \(\Rightarrow...\)

Điều kiện \(\left\{{}\begin{matrix}\dfrac{4x-3x^2y-9xy^2}{x+3y}\ge0\\x+3y\ne0\end{matrix}\right.\)

Với \(3y\ge x\), hệ tương đương:

\(\left\{{}\begin{matrix}\left(x^4-2x^2+4\right)\left(x^2+2\right)=6x^5y\\\left(3y-x\right)^2=\dfrac{4x}{x+3y}-3xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^6+8=6x^5y\left(1\right)\\x^3+27y^3=4x\end{matrix}\right.\left(I\right)\)

Vì \(x=0\) thì hệ vô nghiệm nên \(x\ne0\), khi đó:

\(\left(I\right)\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{8}{x^6}=\dfrac{6y}{x}\\1+\dfrac{27y^3}{x^3}=\dfrac{4}{x^2}\end{matrix}\right.\)

Đặt \(\dfrac{3y}{x}=a,\dfrac{2}{x^2}=b\) ta được hệ:

\(\Leftrightarrow\left\{{}\begin{matrix}1+a^3=2b\\1+b^3=2a\end{matrix}\right.\)

Giải hệ này ta được \(a=b\Leftrightarrow\dfrac{3y}{x}=\dfrac{2}{x^2}\Leftrightarrow y=\dfrac{2}{3x}\)

\(\left(1\right)\Leftrightarrow x^6-4x^4+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=\sqrt{1+\sqrt{5}}\\x=-\sqrt{1+\sqrt{5}}\end{matrix}\right.\)

TH1: \(x=\sqrt{2}\Rightarrow y=\dfrac{\sqrt{2}}{3}\)

TH2: \(x=-\sqrt{2}\Rightarrow y=-\dfrac{\sqrt{2}}{3}\)

TH3: \(x=\sqrt{1+\sqrt{5}}\Rightarrow y=\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)

TH4: \(x=-\sqrt{1+\sqrt{5}}\Rightarrow y=-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)

Đối chiếu với các điều kiện ta được \(\left(x;y\right)=\left(-\sqrt{1+\sqrt{5}};-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\right)\)

ĐKXĐ:...

Từ pt đầu:

\(\Leftrightarrow y^2+y\sqrt{y^2+1}=x-2y+\dfrac{1}{2}\)

\(\Leftrightarrow y^2+1+2y\sqrt{y^2+1}+y^2=2x-4y+2\)

\(\Leftrightarrow\left(\sqrt{y^2+1}+y\right)^2=2x-4y+2\)

\(\Leftrightarrow\sqrt{y^2+1}+y=\sqrt{2x-4y+2}\)

Thế xuống pt dưới:

\(x+\sqrt{x^2-2x+5}=1+2\sqrt{y^2+1}+2y\)

\(\Leftrightarrow\left(x-1\right)+\sqrt{\left(x-1\right)^2+4}=2y+\sqrt{\left(2y\right)^2+4}\)

Do hàm \(t+\sqrt{t^2+4}\) đồng biến

\(\Leftrightarrow x-1=2y\Rightarrow x=2y+1\)

Thế vào pt đầu:

\(\left(y+1\right)^2+y\sqrt{y^2+1}=2y+\dfrac{5}{2}\)

\(\Leftrightarrow y^2+y\sqrt{y^2+1}=\dfrac{3}{2}\)

\(\Leftrightarrow\left(\sqrt{y^2+1}+y\right)^2=4\)

\(\Leftrightarrow\sqrt{y^2+1}+y=2\)

\(\Leftrightarrow\sqrt{y^2+1}=2-y\)

\(\Leftrightarrow...\)